How prove this $\sum_{k=0}^n \binom{n}{k} \binom{(p-1)n}{k} \binom{pn+k}{k} = \binom{pn}{n}^2 $ I think the following equality is true ($p\in \mathbb{N},p\ge 2$):
$$\sum_{k=0}^n \binom{n}{k} \binom{(p-1)n}{k} \binom{pn+k}{k} = \binom{pn}{n}^2 $$
when $p=2$, then
$$\sum_{k=0}^{n}\binom{n}{k}^2\binom{2n+k}{k}=\binom{2n}{n}^2$$
But I can't prove this
and I can't prove this  $p\ge 3$ ?  
Thank you for your help

Suppose we seek to verify that
$$\sum_{k=0}^n {n\choose k}
{pn-n\choose k} {pn+k\choose k} = {pn\choose n}^2.$$
We use the integrals
$${pn-n\choose k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{pn-n}}{z^{k+1}} \; dz$$
and
$${pn+k\choose k} =
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{pn+k}}{w^{k+1}} \; dw.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{pn-n}}{z}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{pn}}{w} 
\sum_{k=0}^n {n\choose k} \frac{(1+w)^k}{z^k w^k}
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{pn-n}}{z}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{pn}}{w} 
\left(1+\frac{1+w}{zw}\right)^n
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{pn-n}}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{pn}}{w^{n+1}} 
(1+w+zw)^n
\; dw \; dz.$$
Expanding the binomial in the inner sum we get
$$\sum_{q=0}^n {n\choose q} w^q (1+z)^q$$
which yields
$$\sum_{q=0}^n {n\choose q}
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{pn-n+q}}{z^{n+1}}
{pn\choose n-q}  \; dz
\\ = \sum_{q=0}^n {n\choose q}
{pn-n+q\choose n}
{pn\choose n-q}.$$
The inner term is
$${n\choose q}
{pn-n+q\choose n}
{pn\choose pn-n+q}
\\ = 
\frac{(pn)!}{q!\times (n-q)! \times
(pn-2n+q)! \times
(n-q)!}
\\ = {pn\choose n}
\frac{n! \times (pn-n)!}{q!\times (n-q)! \times
(pn-2n+q)! \times
(n-q)!}
\\ = {pn\choose n}
{n\choose q} {pn-n\choose n-q}.$$
Thus it remains to show that
$$\sum_{q=0}^n {n\choose q} {pn-n\choose n-q} = {pn\choose n}.$$
This can be done combinatorially or using the integral
$$\frac{1}{2\pi i}
\int_{|v|=\epsilon} \frac{(1+v)^{pn-n}}{v^{n+1}}
\sum_{q=0}^n {n\choose q} v^q \; dv
\\ = \frac{1}{2\pi i}
\int_{|v|=\epsilon} \frac{(1+v)^{pn-n}}{v^{n+1}}
(v+1)^n \; dv
\\ = \frac{1}{2\pi i}
\int_{|v|=\epsilon} \frac{(1+v)^{pn}}{v^{n+1}}
= {pn\choose n}.$$
