Question: Prove using the binomial theorem, that for natural numbers $n$ and $k$ and real positive number $a$, $n^k<(1+a)^n$ for all $n>N$.

Using the binomial theorem, $(1+a)^n=1+na+\dots+\frac{n(n-1)\dots(n-k+1)(n-k)}{(k+1)!}a^{k+1}+\dots+a^n$

For sufficiently large $n$, $\frac{n(n-1)\dots(n-k+1)(n-k)}{(k+1)!}a^{k+1}\sim\frac{n^k(n-k)}{(k+1)!}a^{k+1}$

So the inequality is equivalent to the fact $\frac{n-k}{(k+1)!}a^{k+1}>1$ for sufficiently large $n$, which can be made true by making $n>\frac{(k+1)!}{a^{k+1}}+k$

Is this method of reasoning valid, do I need to be more careful with approximating things in the limit and is there a better way of establishing the inequality still using the binomial theorem (I know $\lim\_{n \to \infty} n^{\frac{1}{n}}=1$ which also proves the inequality).

Answer: 
For there to be an inverse, the original function has to pass the horizontal line test-there must not be two points on the curve with the same $y$ value. Your function satisfies that. A graph (below) is not a proof, but very suggestive. In this case you can prove it by showing $y' \gt 0$, which shows the function is monotonic. If you solve $y=\frac x{1-x^2}$ for $x$ you get $$y=\frac x{1-x^2}\\y-x^2y=x\\x^2y+x-y=0\\x=\frac {-1}{2y} \pm \frac {\sqrt {1+4y^2}}{2y}$$ You need to recognize that you can't divide by zero when $y=0$. We resolve the sign ambiguity by noting that the signs of $x$ and $y$ are always the same. This will be true with the plus sign so the inverse is the unique $$x=\begin {cases} \frac {-1}{2y} + \frac {\sqrt {1+4y^2}}{2y} & y \neq 0\\0&y=0\end {cases}$$ The resulting curve is a nice reflection in $x=y$

[![enter image description here](https://i.stack.imgur.com/jlpuC.jpg)](https://i.stack.imgur.com/jlpuC.jpg)
[![enter image description here](https://i.stack.imgur.com/ltnqL.png)](https://i.stack.imgur.com/ltnqL.png)
Inverse function is 
$$ x = y/(1-y^2)$$
solving for y(x) we can choose any one of the two valid inverse functions
$$ y(x)= -\frac{1}{2x}\pm \sqrt { 1+\frac{1}{(2x)^2} }$$