Inductive proof that ${2n\choose n}=\sum{n\choose i}^2.$

I would like to prove inductively that $${2n\choose n}=\sum_{i=0}^n{n\choose i}^2.$$

I know a couple of non-inductive proofs, but I can't do it this way. The inductive step eludes me. I tried naively things like $${2n+2\choose n+1}={2n+2\over n+1}{2n+1\choose n}=2\cdot {2n+1\over n+1}{2n\choose n},$$

But I don't think it can lead me anywhere. I would like the proof to be as simple as possible.

Solution 1:

Split the $2n$ elements into two groups of size $n$ Then the no. of ways of choosing $n$ from the $2n$ is the no. of ways of choosing $i$ from the 1st and $n-i$ from the 2nd and letting $i$ vary.

Solution 2:

After Martin Sleziak and Marc van Leeuwen's comments, I've found this inductive proof of Vandermonde's identity. (On this very site.)