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?
 A: 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.
A: 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 $j\mapsto n-k-1$
A: $\def\peq{\mathrel{\phantom{=}}{}}$For the first identity, because for $0 \leqslant j \leqslant n - 1$,\begin{align*}
&\peq \binom{n}{j} j (1 - x)^{n - j} x^{j - 1} - \binom{n}{j + 1} (j + 1) (1 - x)^{n - j - 1} x^j\\
&= \frac{n!}{(j - 1)!\, (n - j)!} (1 - x)^{n - j} x^{j - 1} - \frac{n!}{j!\, (n - j - 1)!} (1 - x)^{n - j - 1} x^j\\
&= \frac{n!}{j!\, (n - j)!} (1 - x)^{n - j - 1} x^{j - 1} (j(1 - x) - (n - j)x)\\
&= \binom{n}{j} (1 - x)^{n - j - 1} x^{j - 1} (j - nx),
\end{align*}
so\begin{align*}
&\peq \sum_{j = n - k}^n \binom{n}{j} (1 - x)^{n - j - 1} x^{j - 1} (j - nx)\\
&= nx^{n - 1} + \sum_{j = n - k}^{n - 1} \left( \binom{n}{j} j (1 - x)^{n - j} x^{j - 1} - \binom{n}{j + 1} (j + 1) (1 - x)^{n - j - 1} x^j \right)\\
&= nx^{n - 1} + \left( \binom{n}{n - k} (n - k) (1 - x)^k x^{n - k - 1} - nx^{n - 1} \right)\\
&= \binom{n}{n - k} (n - k) (1 - x)^k x^{n - k - 1}.
\end{align*}
For the second identity, note that\begin{gather*}
\frac{1}{n} \binom{n - 1}{k} \binom{n}{n - k} (n - k) = \frac{1}{n} · \frac{(n - 1)!}{k!\, (n - k - 1)!} · \frac{n!}{k!\, (n - k)!} · (n - k)\\
= \frac{(n - 1)!}{k!\, (n - k - 1)!} · \frac{(n - 1)!}{k!\, (n - k - 1)!} = \left( \binom{n - 1}{k} \right)^2,
\end{gather*}
thus dividing by $x^{n - 1}$ on both sides of the second identity and denoting $m = n - 1$, $t = -\dfrac{1}{x}$, it is equivalent to prove that\begin{gather*}
\sum_{k = 0}^m \left( \binom{m}{k} \right)^2 (t + 1)^k = \sum_{k = 0}^m \binom{m}{k} \binom{m + k}{k} t^{m - k}.\tag{2$'$}
\end{gather*}
Consider the polynomial $(s + t + 1)^m (s + 1)^m$. Since\begin{gather*}
(s + t + 1)^m (s + 1)^m = \left( \sum_{k = 0}^m \binom{m}{k} (t + 1)^k s^{m - k} \right) \left( \sum_{k = 0}^m \binom{m}{k} s^k \right),\\
(s + t + 1)^m (s + 1)^m = (s + 1)^m \sum_{k = 0}^m \binom{m}{k} (s + 1)^k t^{m - k} = \sum_{k = 0}^m \binom{m}{k} t^{m - k} (s + 1)^{m + k},
\end{gather*}
then\begin{gather*}
\left( \sum_{k = 0}^m \binom{m}{k} (t + 1)^k s^{m - k} \right) \left( \sum_{k = 0}^m \binom{m}{k} s^k \right) = \sum_{k = 0}^m \binom{m}{k} t^{m - k} (s + 1)^{m + k}.\tag{3}
\end{gather*}
Treating $t$ as a constant and $s$ as a variable for the moment, the coefficient of $s^m$ of the LHS of (3) is$$
\sum_{k = 0}^m \binom{m}{k} (t + 1)^k · \binom{m}{k} = \sum_{k = 0}^m \left( \binom{m}{k} \right)^2 (t + 1)^k,
$$
and that of the RHS of (3) is$$
\sum_{k = 0}^m \binom{m}{k} t^{m - k} · \binom{m + k}{m} = \sum_{k = 0}^m \binom{m}{k} \binom{m + k}{k} t^{m - k}.
$$
Therefore (2$'$) is true.
A: The first Identity follows by induction. When $k=0$
\begin{eqnarray*}
\binom{n}{n} \frac{x^{n-1}}{1-x} (n-nx) = \binom{n}{n} nx^{n-1}.
\end{eqnarray*}
Now assume the sum for $k$ and add the $(k+1)^{th}$ term ($j=n-k-1$)
\begin{eqnarray*}
\binom{n}{n-k}(n-k)(1-x)^kx^{n-k-1}+\binom{n}{n-k-1}(1-x)^{k}x^{n-k-2}(n-k-1-nx) = 
\binom{n}{n-k-1}x^{n-k-2}(1-x)^k \left( (k+1)x +(n-k-1)-nx \right) =\binom{n}{n-k-1}(n-k-1)(1-x)^{k+1}x^{n-k-2}.
\end{eqnarray*}
