right handside is the square of C(n.k). separate the factors. C(n,k) * C(n.k)
and from the combinational identity we know that C(n,k) can be rewritten as C(n, n-k)

now using the Theorem C(m+n, k) = Sigma...

OK leT me attach the solution that is faster.
I have attached two photos.

For the first one you should notice that the left hand side looks an awful lot like a derivative. In particular, what happens if you differentiate
\[ (1 + x)^n = \sum_{k=0}^n {{n}\choose{k}} x^k\]
and evaluate it at $x = 1$?