1^infinity=Q

lim(x->infinity)(1+Q/x)^x, Q is some positive real number
direct Substitution yields: (1+Q/infinity)^infinity=(1+0)^infinity=1^infinity
because lim(n->infinity)(1/n)=0
then let some number "S"=lim(x->infinity)(1+Q/x)^x
natural log: lnS=lim(x->infinity)ln(1+Q/x)^x
because compound function limit properties: g(lim(x->infinity)f(x))=lim(x->infinity)g(f(x))
property of natural logs:lnS=lim(x->infinity)x*ln(1+Q/x)
x=1/(1/x):lnS=lim(x->infinity)ln(1+Q/x)/(1/x)
L'Hopital's rule:lnS=lim(x->infinity)(-Q/(x^2+Q*x))/(-1/x^2)
reduce:lnS=lim(x->infinity)(Q*x)/(Q+x)
(brief digression)some expression "A" fulfills the following equation: (Q(Q+x))/(Q+x)+A/(Q+X)=(Q*x)/(Q+x)
multiply by (Q+x) on both sides: Q^2+Q*x+A=Q*x
subtract (Q^2+Q*x) from sides:A=Q*x-Q*x-Q^2
reduce:A=-Q^2
substitution:lnS=lim(x->infinity)(Q(Q+x))/(Q+x)-Q^2/(Q+x)
reduce:lnS=lim(x->infinity)(Q-Q^2/(Q+x))
take limit:lnS=Q-Q^2/(Q+infinity)
reduce:lnS=Q-0=Q
raise to e on both sides: S = Q

I'll explain any part that sounds confusing. (though I'll refer back to previous posts if people ask the same question)

oops, S = e^Q

Doesn't Q just equal 1?

Amazingly, no it does not. Part of this has to do with the fact that infinity isn't a number (it's kinda like i in that it allows for some really strange stuff to occur).

informal proof:
let infinity=inf (for brevity) and "t" equal some real number
1^inf = t
(1+0)^inf = t
a-a=0
(1+a-a)^inf = t
Now try to expand this quantity.
you'll inevitably get things like 6*a^(2*inf) and a^(3*inf)-a^inf and other such nonsense that plays havoc with the exponential rules. After all, the number of numbers between one and one million is much greater than the number of numbers between one and two yet the "amount" of both is still infinity. This then leads to the question: does e^inf = inf? Does 2+inf = inf? How about inf/3 = inf? And because those questions above are equal some of the time but not all the time (lim(x->inf)e^x=lim(x->inf)x but lim(x->inf)2+arctan(x)=/lim(x->inf)arctan(x)), that things like 1^inf = 4 exist. If it makes you feel saner, treat infinity like an unreal number such that inf=inf or inf=/inf. There's an entire branch devoted to this idea of seperate infinities and I think it's called "Aleph numbers" I haven't personally studied it and I invite someone who has to correct me but I think they define aleph one infintity as the number of points from zero to one on a line segement. Aleph two infinity is the number of points from the square created from the corners (0,0),(0,1),(1,0),(1,1). Aleph three infinity is the number of points in the cube with corners (0,0,0),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,0,1),(1,1,0),(1,1,1). Aleph four is the number of points in the previous cube that exists for one unit of time with the numebr of points total equaling the number of points for each instant of existence (some use different fourth dimensions, I prefer time since it's easier to visualize). etc.

"I haven't personally studied it and I invite someone who has to correct me but I think they define aleph one infintity as the number of points from zero to one on a line segement. Aleph two infinity is the number of points from the square created from the corners (0,0),(0,1),(1,0),(1,1). Aleph three infinity is the number of points in the cube with corners"

No. Aleph 0 is the cardinality (size, colloquially) of the natural numbers. Aleph 1 is the least cardinality larger than Aleph 0, and so on. Whether or not the real number line has cardinality equal to Aleph 1 is undecidable. The real "square", "cube", and so on all have the same cardinality as the real number line.

There's a simple proof of that last part: given a pair (a,b) in the square [0,1]x[0,1], we can write "a" as 0.a1a2a3... where a1,a2,a3 are decimal digits. Note that in the case a = 1, we let 9 = a1 = a2 = .... As we all know, 0.999... = 1. Similarly write b = 0.b1b2b3.... Then we can map the square to [0,1] by letting (a,b) map to the element 0.a1b1a2b2a3b3.... This is easily seen to be injective. Hence, the cardinality of the square is no greater than the cardinality of [0,1]. Furthermore, the cardinality of the square is no less than [0,1], because the square contains [0,1] as a subset. Thus the only option is that the cardinality of the square is equal to the cardinality of [0,1]. The proof can easily be extended to show that [0,1]^n for any positive integer n has the same cardinality as [0,1].

1^infinity=(1*1*1*1...) series converges to 1.
x*1=x (multiplication by 12 results in same number)
1^infinity=1

>>6
Not a real number, keep your idiotic trolling to one post please.

>>7
Are you saying multiplication by 1(any amount of times)
Changes the number? This is ridiculous.

>>8
"Infinity" isn't an amount of times, gtfo troll

>>9
Infinity is arbitrary large number.Its infinite amount of times.
No limit.If its isn't amount of times,then you speaking of something else.

>>10
You're shoveling so fast that you can't even keep track of it. Infinity is not an arbitrary large number. If that were the case, we could just substitute in, for example, Graham's number, and then the whole 'undefined' thing would just go away.

Listen,infinity doesn't change logic.Its not important how large it is.
1*1*1*1*1... never changes from 1,no matter how hard you try.
x*1=x continues to hold every time you multiply it by 1,
What is so hard to understand?

>>12
because infinity is not a number and is not a real amount. It's closer to an unreal than a real (because any positive real number times infinity is infinity) and therefore can have certain properties inherent to the real numberline that, well, bend in the complex numberline. I can define an unreal for which the multiplicative property of 1 does not apply. I'll call this unreal "f" such that f*1=/f. When I take 1^f, I'm going to get some wierd answers. Yet just because I've taken a real number to an unreal exponent, doesn't mean I can't have a real answer. E.G. e^(pi*i) = -1. Therefore, 1^infinity can equal 1, but it can also equal 67 and more.

>>13
then 1=67 and every number is equal.

Also if infinity is not real (which it is used for by wikipedia)
All proofs containing expression of form lim(N grows to infinity)
cannot be true,because N cannot grow to Infinity,since its not
IN THE REAL NUMBER SYSTEM.
Either you use it as real number
or lose it.Not third option here.

Analogy lim (N>>Infinity)
lim (N>> Becomes apple)
N does not becomes apple ever.It even can't be close or in the same
frame of reference.

It would except for the fact that infinity/infinity has more than one answer and most of those answers are distinct (one that isn't is lim(x->infinity)ln(x)/x Vs. lim(x->infinity)x/x^2 (both are 0)). Again, infinity does not play by the normal rules. Infinity does not have to equal any ONE thing. Infinity is not part of the standard real numberline. Also, because Infinity equals different things under different circumstances, all reals stay their respective values since no function can have two values as you approach infinity for that function.

Just because you use a non-real "number" in a real number expression, does not automatically force the answer to be non-real. Again, e^(pi*i) = -1. Though any number towards apple does not equal apple and can not ever equal apple. Therefore that function diverges and never converges to any real number.

>>16
Infinity is not a function like (take root of 4,and find all roots)
its an upper bound on numbers.like xFF(255) is most you can stuff into a byte.

Just because you use a non-real "number" in a real number expression
Nothing wrong with that but why mathemathicians forbid it.
Every mathematician  i seen,when he views (1/10)^infinity
declares it null and void because ("Infinity is not a number,and cannot be
used in equations")