Prove an identity involving factorials inside summations Prove that: 
\begin{align*}\sum _{k=0}^N\left(k\cdot \frac{N!}{k!\left(N-k\right)!}\cdot \frac{\left(X+k-1\right)!}{\left(X-1\right)!}\cdot \frac{\left(L-X+N-k-1\right)!}{\left(L-X-1\right)!}\frac{\left(L-1\right)!}{\left(L+N-1\right)!}\right)&= \frac{XN}{L}
\end{align*}
for $[0<X<L]\in \mathbb{N}$
My attempt using induction: (I'll leave out the verification of the base cases, but they do hold). Factoring the non-iterates out and moving them to the other side of the equality of the induction hypothesis:
\begin{align*}
\frac{N!}{\left(X-1\right)!\left(L-X-1\right)!}\sum _{k=0}^N\frac{k}{k!\left(N-k\right)!}\cdot \left(X+k-1\right)!\cdot \left(L-X+N-k-1\right)!&=\frac{XN}{L!}\left(N+L-1\right)!
\end{align*}
From the domain of $X$ and $L, \quad X-1 \geq 0$ and $L-X-1 \geq 0$. The $(X+k-1)!$ term can be written as $(k+X-1)(k+X-2)...(k+1)k!$ -- similarly, the $\left(L-X+N-k-1\right)!$ term is expressible as $(N-k+L-X-1)(N-k+L-X-2)...(N-k+1)(N-k)!$ We can cancel the factorial terms in the denominator of the first term of the summation and express the rising factorials using product notation. This returns the induction hypothesis that will be used.
\begin{align*}
\frac{N!}{(X-1)!(L-X-1)!}\sum_{k=0}^N\left(k\prod_{i=1}^{X-1}(k+i)\prod_{i=1}^{L-X-1}(N-k+i)\right) &= \frac{XN}{L!}\left(N+L-1\right)!
\end{align*}
Inductive step: 
First we note the required form: $\frac{X(N+1)}{L!}(L+N)!$.
Through some algebraic manipulation, we arrive at the following. 
\begin{align*}
&(N+1)\left(\frac{N!}{(X-1)!(L-X-1)!}\sum_{k=0}^{N}\left(k\prod_{i=1}^{X-1}(k+i)\prod_{i=1}^{L-X-1}(N+1-k+i)\right)+\frac{(X+N)!}{(X-1)!}\right)
\end{align*}
Because of the $N+1$ nested in the second product of the summation, there does not seem to be a way to insert the induction hypothesis. If one eliminates the $+1$ via altering the limits of the product, a new $k$ term appears:
Inductive step:
\begin{align*}
&(N+1)\left(\frac{N!}{(X-1)!(L-X-1)!}\sum_{k=0}^{N}\left(k\prod_{i=1}^{X-1}(k+i)\prod_{i=2}^{L-X}(N-k+i)\right)+\frac{(X+N)!}{(X-1)!}\right)
\end{align*}
Compared to the induction hypothesis:
\begin{align*}
\frac{N!}{\left(X-1\right)!\left(L-X-1\right)!}\left(\sum _{k=0}^N\left(k\left(\prod _{i=1}^{X-1}\left(k+i\right)\right)\left(\prod _{i=2}^{L-X}\left(N-k+i\right)\right)\left(\frac{\left(N-k+1\right)}{N-k+L-X}\right)\right)\right)
\end{align*}
Is there any way to resolve this, or is a proof by induction not possible?

Here is an approach without using induction. We start with the left-hand side, do some simplifications to finally obtain the right-hand side.

We obtain
  \begin{align*}
\sum _{k=0}^N& k\cdot \frac{N!}{k!\left(N-k\right)!}\cdot \frac{\left(X+k-1\right)!}{\left(X-1\right)!}\cdot \frac{\left(L-X+N-k-1\right)!}{\left(L-X-1\right)!}\frac{\left(L-1\right)!}{\left(L+N-1\right)!}\\
&=\frac{N!(L-1)!}{(L+N-1)!}\sum_{k=1}^N\frac{(X+k-1)!}{(k-1)!(X-1)!}\cdot\frac{(L-X+N-k-1)!}{(N-k)!(L-X-1)!}\tag{1}\\
&=\binom{L+N-1}{N}^{-1}\sum_{k=1}^N X\binom{X+k-1}{k-1}\binom{L-X+N-k-1}{N-k}\tag{2}\\
&=X\binom{L+N-1}{N}^{-1}\sum_{k=0}^{N-1}\binom{X+k}{k}\binom{L-X+N-k-2}{N-k-1}\tag{3}\\
&=X\binom{L+N-1}{N}^{-1}\sum_{k=0}^{N-1}\binom{-X-1}{k}(-1)^k\binom{-L+X}{N-1-k}(-1)^{N-1-k}\tag{4}\\
&=(-1)^{N-1}X\binom{L+N-1}{N}^{-1}\binom{-L-1}{N-1}\tag{5}\\
&=X\binom{L+N-1}{N}^{-1}\binom{L+N-1}{N-1}\tag{6}\\
&=\frac{XN}{L}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we do some rearrangements, factor out terms independent of the index variable $k$ and start with index $k=1$ since the left-hand expression contains the factor $k=0$.

*In (2) we rewrite the expression using binomial coefficients.

*In (3) we shift the index by one to start with $k=0$.

*In (4) we use the rule $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (5) we apply Vandermonde's identity.

*In (6) we apply again the rule from (4).

Note: A proof by induction is also possible. But, if a direct transformation is within reach, this often provides more insight. The main aspect here which connects the left-hand side and the right-hand side is an application of Vandermonde's identity. We would probably  not see this, when proving the identity using induction.

