Note that
\begin{eqnarray*}
(1+x)^{2n} & = & (1+x)^{n}(1+x)^{n}\\
& = & \left[\sum_{i=0}^{n}\binom{n}{i}x^{i}\right]\left[\sum_{j=0}^{n}\binom{n}{j}x^{j}\right]
\end{eqnarray*}
Expand the right hand side and observe that the $x^{n}$ term is
\begin{eqnarray*}
& & \binom{n}{0}x^{0}\cdot\binom{n}{n}x^{n}+\binom{n}{1}x^{1}\cdot\binom{n}{n-1}x^{n-1}+\ldots+\binom{n}{n-1}x^{n-1}\cdot\binom{n}{1}x^{1}+\binom{n}{n}x^{n}\cdot\binom{n}{0}x^{0}\\
& = & \left[\sum_{i=0}^{n}\binom{n}{i}\binom{n}{n-i}\right]x^{n}\\
& = & \left[\sum_{i=0}^{n}\binom{n}{i}\binom{n}{i}\right]x^{n}.
\end{eqnarray*}
The $x^{n}$ in the left hand side is just $\binom{2n}{n}x^{n}$.
Therefore $\sum_{i=0}^{n}\binom{n}{i}^{2}=\binom{2n}{n}$.