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: 
You want to solve, with respect to $x$, the equation
$$
\frac{x}{1-x^2}=y
$$
with the condition that $-1<x<1$. If $y=0$, then $x=0$, so we can assume $y\ne0$. Thus the equation becomes
$$
x^2+\frac{1}{y}x-1=0
$$
Note that this equation has one negative and one positive solution. From the fact that $1-x^2>0$, we need to choose the positive solution for $y>0$ and the negative one for $y<0$. So we have
$$
x=\begin{cases}
\dfrac{-1/y-\sqrt{1/y^2+4}}{2} & \text{if $y<0$} \\[6px]
0 & \text{if $y=0$} \\[6px]
\dfrac{-1/y+\sqrt{1/y^2+4}}{2} & \text{if $y>0$}
\end{cases}
$$

Here's a diagram showing that the inverse function is correct:

[![enter image description here](https://i.stack.imgur.com/KWBHF.png)](https://i.stack.imgur.com/KWBHF.png)

If you follow the blue branch from $-\infty$ to $0$ and the red branch from $0$ to $\infty$, you get exactly the reflection around the bisector of the first and third quadrant of the graph of $f(x)=x/(1-x^2)$ restricted to $(-1,1)$.

Note that for $y>0$ we have
$$
0<\dfrac{-1/y+\sqrt{1/y^2+4}}{2}<1
$$
because, setting $t=1/y$ this is equivalent to
$$
\begin{cases}
\sqrt{t^2+4}>t\\
\sqrt{t^2+4}<2+t
\end{cases}
$$
The first inequality is obviously true; squaring the second one, we get
$$
t^2+4<4+4t+t^2
$$
which is true as well. Similarly, you can prove that, for $y<0$
$$
-1<\dfrac{-1/y-\sqrt{1/y^2+4}}{2}<0
$$
Inverse function is 
$$ x = y/(1-y^2)$$
solving for y(x) we can choose any one of the two valid inverse functions
$$ y(x)= -\frac{1}{2x}\pm \sqrt { 1+\frac{1}{(2x)^2} }$$