Inductive proof for $\binom{2n}{n}=\sum\limits_{k=0}^n\binom{n}{k}^2$ I want to prove the following identity using induction (not double counting method). Although it is a specific version of Vandermonde's identity and its inductive proof is presented here, but I need a direct inductive proof on this, not the general form.
$$\binom{2n}{n}=\sum\limits_{k=0}^n\binom{n}{k}^2$$
I have tried to simplify $\binom{2n+2}{n+1}$ using Pascal's theorem, but did not get any result. Any help?
 A: In general, one can show by induction that
$$\frac{d^n}{dt^n} f(t)g(t) = \sum_{k=0}^n \binom{n}{k} \biggl(\frac{d^k}{dt^k}f(t) \biggr) \biggl(\frac{d^{n-k}}{dt^{n-k}}g(t) \biggr). \tag{1} $$
For $0 \le k \le r$ integers, one can see by induction that
$$\frac{d^k}{dt^k} t^r = r(r-1)\cdots(r-k+1)t^{r-k} = k!\binom{r}{k}t^{r-k}. $$
So in particular,
$$\frac{d^n}{dt^n} t^{2n} = n!\binom{2n}{n}t^n. \tag{2} $$
Applying $(1)$ to $f(t) = g(t) = t^n$ gives
\begin{align}
\frac{d^n}{dt^n} t^{2n}
&= \sum_{k=0}^n \binom{n}{k} \biggl(\frac{d^k}{dt^k}t^{n} \biggr) \biggl(\frac{d^{n-k}}{dt^{n-k}}t^{n} \biggr) \\
&= \sum_{k=0}^n \binom{n}{k} \cdot k!\binom{n}{k}t^{n-k} \cdot (n-k)!\binom{n}{n-k} t^k \\
&= \sum_{k=0}^n n! \binom{n}{k}^2 t^n. \tag{3}
\end{align}
Comparing $(2)$ and $(3)$ gives the desired result.
A: Suppose
$$
\sum_{k=0}^n\binom{n}{k}^2=\binom{2n}{n}\tag{1}
$$
then
$$
\begin{align}
\sum_{k=0}^{n+1}\binom{n+1}{k}^2
&=\sum_{k=0}^{n+1}\left[\binom{n}{k}+\binom{n}{k-1}\right]^2\tag{2a}\\
&=\sum_{k=0}^{n+1}\left[\binom{n}{k}^2+\binom{n}{k-1}^2+2\binom{n}{k}\binom{n}{k-1}\right]\tag{2b}\\
&=\binom{2n}{n}\left(1+1+\frac{2n}{n+1}\right)\tag{2c}\\
&=\binom{2n+2}{n+1}\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a)}$: Pascal's Triangle identity
$\text{(2b)}$: algebra
$\text{(2c)}$: apply $(1)$ and $(3)$
$\text{(2d)}$: $\binom{2n+2}{n+1}=\frac{4n+2}{n+1}\binom{2n}{n}$

Lemma:
$$
\sum_{k=1}^n\binom{n}{k}\binom{n}{k-1}=\frac{n}{n+1}\sum_{k=0}^n\binom{n}{k}^2\tag{3}
$$
Proof:
Since $\binom{n}{k-1}=\frac{k}{n-k+1}\binom{n}{k}$, we have $\binom{n}{k}+\binom{n}{k-1}=\frac{n+1}{n-k+1}\binom{n}{k}$. Therefore,
$$
\frac{n-k+1}{n+1}\left[\binom{n}{k}+\binom{n}{k-1}\right]\binom{n}{k-1}=\binom{n}{k}\binom{n}{k-1}\tag{3a}
$$
Since $\binom{n}{k}=\frac{n-k+1}{k}\binom{n}{k-1}$, we have $\binom{n}{k}+\binom{n}{k-1}=\frac{n+1}{k}\binom{n}{k-1}$. Therefore,
$$
\frac{k}{n+1}\left[\binom{n}{k-1}+\binom{n}{k}\right]\binom{n}{k}=\binom{n}{k-1}\binom{n}{k}\tag{3b}
$$
Adding $(3a)$ and $(3b)$ and cancelling yields
$$
\frac{n-k+1}{n+1}\binom{n}{k-1}^2+\frac{k}{n+1}\binom{n}{k}^2=\binom{n}{k-1}\binom{n}{k}\tag{3c}
$$
Summing $(3c)$ over $k$, and substituting $k\mapsto k+1$ in the leftmost sum, gives
$$
\frac{n}{n+1}\sum_{k=0}^n\binom{n}{k}^2=\sum_{k=1}^n\binom{n}{k-1}\binom{n}{k}\tag{3d}
$$
QED
