<s>[INST] 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?
 [/INST] First Identity
$$
\begin{align}
&\sum_{j=n-k}^n\binom{n}{j}(1-x)^{n-j-1}x^{j-1}(j-nx)\\
&=\sum_{j=n-k}^n\binom{n}{j}(1-x)^{n-j-1}x^{j-1}[j(1-x)-(n-j)x]\tag1\\
&=\sum_{j=n-k}^nn\binom{n-1}{j-1}(1-x)^{n-j}x^{j-1}-\sum_{j=n-k}^nn\binom{n-1}{j}(1-x)^{n-j-1}x^{j}\tag2\\
&=\sum_{j=n-k-1}^{n-1}n\binom{n-1}{j}(1-x)^{n-j-1}x^{j}-\sum_{j=n-k}^nn\binom{n-1}{j}(1-x)^{n-j-1}x^{j}\tag3\\[3pt]
&=n\binom{n-1}{n-k-1}(1-x)^{k}x^{n-k-1}\tag4\\[9pt]
&=(n-k)\binom{n}{n-k}(1-x)^{k}x^{n-k-1}\tag5
\end{align}
$$
Explanation:
$(1)$: rewrite $j-nx=j(1-x)-(n-j)x$
$(2)$: $\binom{n}{j}j=\binom{n-1}{j-1}n$ and $\binom{n}{j}(n-j)=\binom{n-1}{j}n$
$(3)$: substitute $j\mapsto j+1$ in the left sum
$(4)$: combine the cancelling terms
$(5)$: $\binom{n-1}{n-k-1}n=\binom{n}{n-k}(n-k)$

Second Identity
$$
\begin{align}
&\sum_{k=0}^{n-1}\frac{(-1)^k}n\binom{n-1}k\binom{n}{n-k}(n-k)(1-x)^kx^{n-k-1}\\
&=\sum_{k=0}^{n-1}(-1)^k\binom{n-1}k\binom{n-1}{n-k-1}(1-x)^kx^{n-k-1}\tag6\\
&=\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}(-1)^k\binom{n-1}{k}\binom{n-1}{n-k-1}\binom{k}{j}(-x)^{k-j}x^{n-k-1}\tag7\\
&=\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}(-1)^j\binom{n-1}j\binom{n-1}{n-k-1}\binom{n-j-1}{k-j}x^{n-j-1}\tag8\\
&=\sum_{j=0}^{n-1}(-1)^j\binom{n-1}{j}\binom{2n-j-2}{n-j-1}x^{n-j-1}\tag9\\
&=(-1)^{n-1}\sum_{k=0}^{n-1}\binom{n-1}{k}\binom{n+k-1}{k}(-x)^k\tag{10}
\end{align}
$$
Explanation:
$\phantom{1}(6)$: $\binom{n}{n-k}\frac{n-k}n=\binom{n-1}{n-k-1}$
$\phantom{1}(7)$: Binomial Theorem: $(1-x)^k=\sum_{j=0}^{n-1}\binom{k}{j}(-x)^{k-j}$
$\phantom{1}(8)$: $\binom{n-1}{k}\binom{k}{j}=\binom{n-1}{j}\binom{n-j-1}{k-j}$
$\phantom{1}(9)$: Vandermonde Identity: $\sum_{k=0}^{n-1}\binom{n-1}{n-k-1}\binom{n-j-1}{k-j}=\binom{2n-j-2}{n-j-1}$
$(10)$: substitute $j\mapsto n-k-1$
</s>