Simplifying sum with rising and falling factorials

In the first sum, the coefficient of $\frac{\gamma^m}{m+1}$ is $$ \sum_{n+p+q=m}\sum_{ l =0}^{n} \frac{(-1)^l\left(q + l \right)!\left(p + n - l\right)!}{p! q!} \frac{ \left(N\right)^{(p)} \left(N\right)_{(q)}}{\left(M + 1\right)^{(p + n - l)}\left( M - 1 \right)_{(q + l)}}. $$ We set $k=q+l$; as $l$ goes from $0$ to $n$, $k$ goes from $q$ to $q+n = m-p$: $$ \sum_{n+p+q=m}\sum_{k=q}^{m-p} \frac{(-1)^{k-q} k!\left(m-k\right)!}{p! q!} \frac{ \left(N\right)^{(p)} \left(N\right)_{(q)}}{\left(M + 1\right)^{(m-k)}\left( M - 1 \right)_{(k)}}. $$ Now we rearrange this sum to depend on $k$ first. For fixed $m$ and $k$, our constraints are that $q \le k$, $p \le m-k$, and $n = m-p-q$. So we have $$ \sum_{k=0}^m \sum_{q=0}^k \sum_{p=0}^{m-k} \frac{(-1)^{k-q} k!\left(m-k\right)!}{p! q!} \frac{ \left(N\right)^{(p)} \left(N\right)_{(q)}}{\left(M + 1\right)^{(m-k)}\left( M - 1 \right)_{(k)}}. $$ Looking only at the $k^{\text{th}}$ term of this sum, we can factor it as $$ \frac{(-1)^k k! (m-k)!}{\left(M + 1\right)^{(m - k)}\left(M - 1\right)_{(k)}} \left(\sum_{q=0}^k (-1)^q \frac{(N)_{(q)}}{q!}\right) \left(\sum_{p=0}^{m-k} \frac{(N)^{(p)}}{p!}\right). $$ The sum over $p$ is known to simplify to $\frac{(N+1)^{(m-k)}}{(m-k)!}$. The sum over $q$ is nearly as well-behaved: we can rewrite $(-1)^q \frac{(N)_{(q)}}{q!}$ as $\frac{(-N)^{(q)}}{q!}$ and then simplify the sum to $\frac{(-N+1)^{(k)}}{k!}$ or $(-1)^k \frac{(N-1)_{(k)}}{k!}$.

These results cancel with the factors that only depend on $k$, so the $k^{\text{th}}$ term of our sum simplifies to $$ \frac{\left(N + 1\right)^{(m - k)}\left( N - 1\right)_{(k)}}{\left(M + 1\right)^{(m - k)}\left(M - 1\right)_{(k)}}. $$ Remembering that we are summing this as $k$ goes from $0$ to $m$, we get exactly the coefficient of $\frac{\gamma^m}{m+1}$ in the sum we wanted to get.