We are showing that when $\alpha$ and $p$ are real and $p>0$ then $\lim_{n\rightarrow\infty}\frac{n^\alpha}{(1+p)^n}=0$
Proof. Let $k$ be an integer such that $k>0$, $k>\alpha$. Then for $n>2k$,$(1+p)^n> \binom{n}{k}p^k=\frac{n(n-1)\cdots(n-k+1)}{k!}p^k>\frac{n^k p^k}{2^k k!} $ Hence for $n>2k$, $0<\frac{n^\alpha}{(1+p)^n}<\frac{2^k k!}{p^k}n^{\alpha-k}$
Since $\alpha-k<0$, $n^{\alpha-k}\rightarrow 0$.
I can follow the whole thing except for where the heck $2^k$ came from. It's clear that it has to do with the fact that the numerator of the big fraction is a product of $k$ terms involving $n$ and $n>2k$ but that's as far as I can see.