Let $k \in N, x \gt 0$. Show that there exists some $n_2 \in \mathbb{N}$ so that $\forall n \geq n_2: (1+x)^n \gt n^k$. Hint: binomial theorem.
My thought on this is first to make the substitution $(1+x)=b$ which means $b>1$ and $b^n>1$. This would also be true if $k=0$ and $n=1$ thus $n_2=1$.
Next step I can think of is using archimedian property $a>0, y \in \mathbb{R}, m \in \mathbb{N}, ma>y$. This will result in $b^n=(1+x)^n=ma > y = n^k$. My current idea would be to replace $n$ with some other cleverly devised number OR using induction because of the request for all $n$ and maybe then I can use binomial theorem to finally solve this. What's really throwing me off is that $n$ is the exponent on the left and also the base on the right (that is why I thought about replacing $n$).
Any hints on how to get to the next step? Thanks.