How can it be shown that $\sum_{k=0}^{n}\binom{n}{k}\binom{2n-k}{n}\left(a-1\right)^k=\sum_{k=0}^{n}\binom{n}{k}^{2}a^{n-k}$ How can it be shown that:

$$\sum_{k=0}^{n}\binom{n}{k}\binom{2n-k}{n}\left(a-1\right)^k=\sum_{k=0}^{n}\binom{n}{k}^{2}a^{n-k}$$


My try:
$$\sum_{k=0}^{n}\binom{n}{k}\binom{2n-k}{n}\left(a-1\right)^k=\sum_{k=0}^{n}\binom{n}{k}\binom{2n-k}{n}\sum_{j=0}^{k}\binom{k}{j}a^{j}\left(-1\right)^{k-j}$$$$=\left(-1\right)^{n}\sum_{k=0}^{n}\binom{n}{k}\binom{-n-1}{n-k}\sum_{j=0}^{k}\binom{k}{j}a^{j}\left(-1\right)^{-j}$$
$$=\left(-1\right)^{n}\sum_{j=0}^{n}\binom{n}{j}a^{j}\left(-1\right)^{-j}\sum_{k=j}^{n}\binom{n-j}{k-j}\binom{-n-1}{n-k}$$
$$=\left(-1\right)^{n}\sum_{j=0}^{n}\binom{n}{j}a^{j}\left(-1\right)^{-j}\binom{-j-1}{n-j}$$
$$=\sum_{j=0}^{n}\binom{n}{j}a^{j}\binom{n}{j}=\sum_{\color{red}{j}=0}^{n}\binom{n}{\color{red}{j}}^2a^{n-\color{red}{j}}$$
The problem is that I have $\color{red}{j}$ instead of $k$.

Source :math.wvu.edu
 A: Replace $j$ by $k$ to get the desired result. Calling the summation index $j$ or $k$ does not make a difference.
A: Here is a slightly different approach, for variety.
We seek to show that
$$\sum_{k=0}^n {n\choose k} {2n-k\choose n} (a-1)^k
= \sum_{k=0}^n {n\choose k}^2 a^{n-k}.$$
Note that with $0\le q\le n$ the coefficients on $[a^q]$ must be equal,
hence we require
$$\sum_{k=q}^n {n\choose k} {2n-k\choose n} {k\choose q} (-1)^{k-q}
= {n\choose q}^2.$$
Starting with the left we note that
$${n\choose k} {k\choose q} =
\frac{n!}{(n-k)! \times q! \times (k-q)!}
= {n\choose q} {n-q\choose n-k}.$$
We are now reduced to showing that
$$\sum_{k=q}^n  {2n-k\choose n} {n-q\choose n-k} (-1)^{k-q}
= {n\choose q}.$$
Starting with the LHS we find
$$\sum_{k=q}^n  {2n-k\choose n-k} {n-q\choose n-k} (-1)^{k-q}
\\ = [z^n] (1+z)^{2n}
\sum_{k=q}^n \frac{z^k}{(1+z)^k} {n-q\choose n-k} (-1)^{k-q}
\\ = [z^n] (1+z)^{2n}  \frac{z^q}{(1+z)^q}
\sum_{k=0}^{n-q} \frac{z^k}{(1+z)^k} {n-q\choose n-q-k} (-1)^{k}
\\ = [z^n] (1+z)^{2n}  \frac{z^q}{(1+z)^q}
\left(1-\frac{z}{1+z}\right)^{n-q}
\\ = [z^n] (1+z)^{2n}  \frac{z^q}{(1+z)^q}
\frac{1}{(1+z)^{n-q}}
= [z^{n-q}] (1+z)^n = {n\choose q}.$$
We obtain the RHS and this concludes the argument.
