How do I show that $\sum_{k = 0}^n \binom nk^2 = \binom {2n}n$? $$\sum_{k = 0}^n \binom nk^2 = \binom {2n}n$$
I know how to "prove" it by interpretation (using the definition of binomial coefficients), but how do I actually prove it?

To prove binomial identities it is often convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$.  So, we can write e.g.
\begin{align*}
\binom{n}{k} = [x^k](1+x)^n\tag{1}
\end{align*}
Note,  this is essentially the same approach as that of @Batominovski but with a somewhat more algebraic look and feel.

We obtain
  \begin{align*}
\sum_{k=0}^{n}\binom{n}{k}^2&=\sum_{k=0}^{n}[x^k](1+x)^n[t^k](1+t)^n\tag{2}\\
&=[x^0](1+x)^n\sum_{k=0}^{n}x^{-k}[t^k](1+t)^n\tag{3}\\
&=[x^0](1+x)^n\left(1+\frac{1}{x}\right)^n\tag{4}\\
&=[x^n](1+x)^{2n}\tag{5}\\
&=\binom{2n}{n}
\end{align*}

Comment:


*

*In (2) we use the coefficient of operator twice according to (1)

*In (3) we use the linearity of the coefficient of operator and the rule $$[x^n]x^{-k}P(x)=[x^{n+k}]P(x)$$

*In (4) we use the substitution rule and apply $x\rightarrow \frac{1}{x}$
\begin{align*}
P(x)&=\sum_{k=0}^na_kx^k=\sum_{k=0}^nx^k[t^k]P(t)\\
P\left(\frac{1}{x}\right)&=\sum_{k=0}^na_kx^{-k}=\sum_{k=0}^nx^{-k}[t^k]P(t)
\end{align*}

*In (5) we use $\left(1+\frac{1}{x}\right)^n=x^{-n}(1+x)^n$ and $[x^0]x^{-n}=[x^n]$
