Proving a combinatorial identity involving sum and integral I want to prove  
Therefore I use that the derivatives of both sides ($\frac{d}{\text{d}p}$) are equal (and that for a fixed p the values of both sides are equal).
Has anyone got a clue how to find that the derivative of the left side is equal to $-\frac{n!}{k!(n-k-1)!}p^k(1-p)^{n-k-1}$? I have been trying hard but I cannot figure it out.
Thanks in advance.
Manuel
 A: Taking the derivative of left member, we have
\begin{align*}
  & \sum_{j=1}^k \binom nj jp^{j-1}(1-p)^{n-j} - \sum_{j=0}^k \binom nj p^j(n-j)(1-p)^{n-j-1} \\
=& \sum_{j=0}^{k-1} \binom n{j+1} (j+1)p^{j}(1-p)^{n-j-1} - \sum_{j=0}^k \binom nj(n-j) p^j(1-p)^{n-j-1}
\end{align*}
Now note that
$$(j+1)\binom n{j+1} = (j+1)\frac{n!}{(j+1)!(n-j-1)!} = (n-j)\frac{n!}{j!(n-j)!} = (n-j)\binom nj$$
So all terms on the LHS sum cancel with the corresponding term in the RHS sum, and only the one you're looking for remains.
A: Using the product rule to take the derivative on the left hand side gives
$$
\begin{aligned}
&\sum_{j=0}^k\left[n\binom{n-1}{j-1}p^{j-1}(1-p)^{n-j}-n\binom{n-1}{j}p^j(1-p)^{n-j-1}\right]\\
&\quad=n\sum_{j=1}^k\binom{n-1}{j-1}p^{j-1}(1-p)^{n-j}-n\sum_{j=1}^{k+1}\binom{n-1}{j-1}p^{j-1}(1-p)^{n-j}\\
&\quad=-n\binom{n-1}{k}p^k(1-p)^{n-k-1}.
\end{aligned}
$$
In the first line, the factors stemming from the exponents that dropped down in taking the derivative have been combined with the binomial coefficients. In the second line, we have noticed that the $j=0$ term in the first sum is zero and have shifted the summation index in the second term. In the third line, we have observed that the sum telescopes.
