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)

>>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.