summation - Show that $sum_{k=0}^{n}binom{n}{k}^2=frac{n+1}{n}sum_{k=1}^{n}binom{n}{k}binom{n}{k-1}$

Show that $\sum_{k=0}^{n}\binom{n}{k}^2=\frac{n+1}{n}\sum_{k=1}^{n}\binom{n}{k}\binom{n}{k-1}$

I came across this result
while trying to solve this:

inductive proof for $\binom{2n}{n}$

My proof is cumbersome,
so I hope that
someone can come up
with a more elegant proof.

Note:
I know that
$\sum_{k=0}^{n}\binom{n}{k}^2 =\binom{2n}{n}$.

Answer

The fact that $$\sum_{k=0}^\infty {n \choose k}^2 = {2n \choose n}$$
has a combinatorial interpretation: to select $n$ items from $2n$, first take an arbitrary subset of the first $n$ items, and if this had cardinality $k$ select $n-k$ of the second $n$ items.

Similarly,

$${2n \choose n} = \sum_{k=1}^{n} {n+1 \choose k} {n-1 \choose n-k}$$

and
$${n+1 \choose k} {n-1 \choose n-k} = \frac{n+1}{n} {n \choose k}{n \choose k-1}$$