Within another answer to a question concerning a sums of the type
$$\sum_{k=0}^n \binom{n}{k}^2$$
there was a simple indetity given which reduces this sum to a simple binomial coefficient, to be exact to
$$\binom{2n}{n}$$
However I tried to prove the formula
$$\sum_{k=0}^n \binom{n}{k}^2~=~\binom{2n}{n}$$
by induction and failed. Overall my attempt was to split up the sum for $n=m+1$ as follows
$$\sum_{k=0}^{m+1} \binom{m+1}{k}^2~=~\sum_{k=0}^m \binom{m+1}{k}^2 + \binom{m+1}{m+1}^2~=~(m+1)^2\sum_{k=0}^m \frac1{(m+1-k)^2}\binom{m}{k}^2 +1$$
I now stuck at reshaping this sum in the end and to substitute it by
$$\binom{2m}{m}$$
so I guess either induction is not the right way to approach to this proof or I made a critical error somewhere. I would be interested in a right proof and a explanation why my attempt did not worked out.
Thank you in advance.
I just figured out this formula is called Vandermondes Identity and it can be derived in way from the Binomial Theorem. But I am still interested in a proof maybe without this property.