How to prove $\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6$? I'd like to find out why
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6
\end{align}
I tried to rewrite it into a geometric series
\begin{align}
\sum_{n=0}^{\infty} \frac{n^2}{2^n} = \sum_{n=0}^{\infty} \Big(\frac{1}{2}\Big)^nn^2
\end{align}
But I don't know what to do with the $n^2$. 

Let me show you a slightly different approach; this approach is very powerful, and can be used to compute values for a large number of series.
Let's think of this in terms of power series.  You noticed that you can write
$$
\sum_{n=0}^{\infty}\frac{n^2}{2^n}=\sum_{n=0}^{\infty}n^2\left(\frac{1}{2}\right)^n;
$$
so, let's consider the power series
$$
f(x)=\sum_{n=0}^{\infty}n^2x^n.
$$
If we can find a simpler expression for the function $f(x)$, and if $\frac{1}{2}$ lies within its interval of convergence, then your series is exactly $f(\frac{1}{2})$.
Now, we note that
$$
f(x)=\underbrace{0}_{n=0}+\sum_{n=1}^{\infty}n^2x^n=x\sum_{n=1}^{\infty}n^2x^{n-1}.
$$
But, notice that $\int n^2x^{n-1}\,dx=nx^n$; so, 
$$\tag{1}
\sum_{n=1}^{\infty}n^2 x^{n-1}=\frac{d}{dx}\left[\sum_{n=0}^{\infty}nx^n\right].
$$
Now, we write
$$\tag{2}
\sum_{n=0}^{\infty}nx^n=\underbrace{0}_{n=0}+x\sum_{n=1}^{\infty}nx^{n-1}=x\frac{d}{dx}\left[\sum_{n=0}^{\infty}x^n\right].
$$
But, this last is a geometric series; so, as long as $\lvert x\rvert<1$,
$$
\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}.
$$
Plugging this back in to (2), we find that for $\lvert x\rvert<1$,
$$
\sum_{n=0}^{\infty}nx^n=x\frac{d}{dx}\left[\frac{1}{1-x}\right]=x\cdot\frac{1}{(1-x)^2}=\frac{x}{(1-x)^2}.
$$
But, plugging this back in to (1), we find that
$$
\sum_{n=1}^{\infty}n^2x^{n-1}=\frac{d}{dx}\left[\frac{x}{(1-x)^2}\right]=\frac{x+1}{(1-x)^3}
$$
So, finally,
$$
f(x)=x\cdot\frac{x+1}{(1-x)^3}=\frac{x(x+1)}{(1-x)^3}.
$$
Plugging in $x=\frac{1}{2}$, we find
$$
\sum_{n=0}^{\infty}\frac{n^2}{2^n}=f\left(\frac{1}{2}\right)=6.
$$
