Proving two binomial identities I would like to show that
\begin{align}
&\sum_{j=n-k}^n\binom nj(1-x)^{n-j-1}x^{j-1}(j-nx)\\
&\qquad=\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\sum_{k=0}^{n-1}\frac{(-1)^k}n\binom{n-1}k\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\\
&\qquad=(-1)^{n-1}\sum_{k=0}^{n-1}\binom{n-1}k\binom{n+k-1}k(-x)^k
\end{align}
I feel I exhausted all identities/properties of binomials without success. Mathematica says it is true, but how to show it?

For the second one we get
$$\sum_{k=0}^{n-1} (-1)^{n-k-1}
\frac{k+1}{n} {n-1\choose k} {n\choose k+1}
x^k (1-x)^{n-1-k}
\\ = (-1)^{n-1} \sum_{k=0}^{n-1} {n-1\choose k} {n+k-1\choose k}
(-1)^k x^k.$$
The LHS is
$$\sum_{k=0}^{n-1} (-1)^{n-k-1}
{n-1\choose k} {n-1\choose k}
x^k (1-x)^{n-1-k}.$$
The coefficient on $[x^q]$ where $0\le q\le n-1$  of the LHS is
$$\sum_{k=0}^{q} (-1)^{n-k-1}
{n-1\choose k} {n-1\choose k}
[x^{q-k}] (1-x)^{n-1-k}
\\ = \sum_{k=0}^{q} (-1)^{n-k-1}
{n-1\choose k} {n-1\choose k}
(-1)^{q-k} {n-1-k\choose q-k}.$$
Using the RHS we therefore must show that
$${n-1\choose q} {n+q-1\choose q}
= \sum_{k=0}^{q}
{n-1\choose k} {n-1\choose k}
{n-1-k\choose q-k}.$$
Now note that
$${n-1\choose k} {n-1-k\choose q-k} =
\frac{(n-1)!}{k! \times (q-k)! \times (n-1-q)!}
= {n-1\choose q} {q\choose k}.$$
This reduces the claim to
$${n+q-1\choose q} =
\sum_{k=0}^q {n-1\choose k} {q\choose k}$$
which is
$$\sum_{k=0}^q {n-1\choose k} {q\choose q-k}
= [z^q] (1+z)^q \sum_{k=0}^q {n-1\choose k} z^k.$$
Here the coefficient extractor enforces the range and we get
$$[z^q] (1+z)^q \sum_{k\ge 0} {n-1\choose k} z^k
= [z^q] (1+z)^q (1+z)^{n-1} \\ = [z^q] (1+z)^{n+q-1}
= {n+q-1\choose q}$$
as required. This last step can also be done by Vandermonde.
