<s>[INST] How can I prove this combinatorial identity without using Wilf-Zeilberger? I am trying to prove the following identity without using W-Z algorithm:
\begin{equation}
\sum\limits_{j=0}^n (-1)^{(n-j)} \frac{(n+j+1)!}{(n-j)!(j)!(j+1)!} = 1
\end{equation}
 [/INST] As observed in the comments, your sum is equal to 
$$
\sum_{j=0}^n (-1)^{n-j} \binom{n}{n-j} \binom{n+j+1}{j+1} \, .
$$
Given a (formal) power series $f(x)$, let $[x^m] f(x)$ denote the coefficient of $x^m$.  Note that
$$
(1 - x)^n = \sum_{j \geq 0} (-1)^j \binom{n}{j} x^j
$$
and
$$
\frac{1}{(1 - x)^{n+1}} = \sum_{j \geq 0} \binom{n+j}{j} x^j \, .
$$
Then
$$
\sum_{j \geq 0} \binom{n+j+1}{j+1} x^j = \frac{1}{x}\left(\frac{1}{(1 - x)^{n+1}} - 1\right) \, .
$$
By the convolution formula for ordinary generating functions, then
\begin{align*}
\sum_{j=0}^n (-1)^{n-j} \binom{n}{n-j} \binom{n+j+1}{j+1} &= [x^n] \left(\sum_{j \geq 0} (-1)^j \binom{n}{j} x^j \right)\left(\sum_{j \geq 0} \binom{n+j+1}{j+1} x^j\right)\\
&= [x^n] (1 - x)^n \cdot \frac{1}{x}\left(\frac{1}{(1 - x)^{n+1}} - 1\right)\\
&= [x^{n+1}]  \left(\frac{1}{1 - x} - (1 - x)^n \right) = 1 - (-1)^{n+1} \binom{n}{n+1}\\
&= 1
\end{align*}
as desired.
</s>