Proving $ \binom n 0 ^2 + \binom n 1 ^2 + \dots + \binom n n ^2 = \binom { 2n} n $ without induction I have to prove that:
$$ \binom n 0 ^2  + \binom n 1 ^2  + \dots + \binom n n ^2 = \binom { 2n} n $$
I don't want a complete solution, but only a hint.

The secret is to expand $(1+x)^{2n}$ in two ways:
$$
\sum_{k=0}^{2n}\binom{2n}{k}x^k=(1+x)^{2n}=\left(\sum_{k=0}^{n}\binom{n}{k}x^k\right)\left(
\sum_{k=0}^{n}\binom{n}{k}x^{n-k}\right).
$$
In the left sum the coefficient of $x^n$ is $\binom{2n}{n}$ while the the right one in
$$
\sum_{k=0}^n\binom{n}{k}\binom{n}{k}=\sum_{k=0}^n\binom{n}{k}^2.
$$
Hence
$$
\binom{2n}{n}=\sum_{k=0}^n\binom{n}{k}^2
$$
Generalization. Similiarly one can obtain that $\,\,\,\displaystyle
\binom{mn}{n}=\sum_{k_1+\cdots+k_m=n}\binom{n}{k_1}\cdots\binom{n}{k_m}.
$
