Show that c(n, 0)*c(n, 1) + c(n, 1)*c(n, 2) + ... + c(n, n - 1)*c(n, n) = (2n)!/[(n - 1)!(n + 1)!]?

Vandermonde's Identity states that
C(m+n,r) = sum(k = 0 to r) C(m, k) * C(n, r-k).

Letting m = n and r = n+1:
C(2n, n+1) = sum(k = 0 to r) C(n, k) * C(n, n+1-k).
==> (2n)! / [(n+1)!(n-1)!] = sum(k = 0 to r) C(n, k) * C(n, n - (n+1-k))
since C(n,k) = C(n,n-k)

==> (2n)! / [(n+1)!(n-1)!] = sum(k = 0 to r) C(n, k) * C(n, k-1)).

I hope this helps!

First off, notice that (2n)!/[(n - 1)!(n + 1)!] = C(2n, n - 1).

Vandermonde's identity, in general, asserts that C(m + n, r) = ∑[k: 0, r] C(m, k)C(n, r - k). Here, we will only use the specific instance:

C(2n, n - 1) = ∑[k: 0, n-1] C(n, k)C(n, (n - 1) - k)

C is symmetric in that C(n, k) = C(n, n - k) (this follows from the definition, you can check easily). From that, we may rewrite C(n, (n - 1) - k) = C(n, n - (k+1)) as C(n, k+1). Hence, Vandermonde's identity asserts that:

C(2n, n - 1) = ∑[k: 0, n - 1] C(n, k)C(n, k+1)

which is precisely the identity we set out to prove!

Note: kb's proof is equally good. Even though he used C(2n, n+1) and I used C(2n, n-1), we of course know that C(2n, n-1) = C(2n, 2n-(n-1)) = C(2n, n+1).