How many ways can you pick $k$ balls from a set of $n$ different red balls and $m$ green different balls?
Answer
$$(n+m)^{\underline k}$$
But you can count them another way. First suppose that the $k$ balls are red, then $k-1$ are red and $1$ is green, etc.
This gives
$$\sum_{j=0}^k\binom kj n^{\underline j} m^{\underline {k-j}}$$
I like Knuth's notation for falling factorials better:
$$
\alpha^{\underline{h}}
= \alpha \cdot (\alpha - 1) \cdot \dotsm \cdot (\alpha - h + 1)
$$
First note that:
$$
\Delta \alpha^{\underline{h}}
= (\alpha + 1)^{\underline{h}} -\alpha^{\underline{h}}
= (\alpha + 1) \cdot \alpha^{\underline{h - 1}}
- \alpha^{\underline{h - 1}} \cdot (\alpha - h + 1)
= h \alpha^{\underline{h - 1}}
$$
This suggests:
$$
\sum_{0 \le k \le n} (\alpha + k)^{\underline{h}}
= \frac{(\alpha + n + 1)^{\underline{h + 1}}}{h + 1}
- \frac{\alpha^{\underline{h + 1}}}{h + 1}
$$
This we prove by induction.
Base: $n = 0$ gives:
$\begin{align}
\sum_{0 \le k \le 0} (\alpha + k)^{\underline{h}}
&= \alpha^{\underline{h}} \\
\frac{(\alpha + 1)^{\underline{h + 1}}}{h + 1}
- \frac{\alpha^{\underline{h + 1}}}{h + 1}
&= \frac{1}{h + 1}
\Delta \alpha^{\underline{h + 1}} \\
&= \alpha^{\underline{h}}
\end{align}$
This checks out.
Induction: Assume it is true for $n$, look at $n + 1$:
$\begin{align}
\sum_{0 \le k \le n + 1} (\alpha + k)^{\underline{h}}
&= \sum_{0 \le k \le n} (\alpha + k)^{\underline{h}}
+ (\alpha + n + 1)^{\underline{h}} \\
&= \frac{(\alpha + n + 1)^{\underline{h + 1}}}{h + 1}
- \frac{\alpha^{\underline{h + 1}}}{h + 1}
+ (\alpha + n + 1)^{\underline{h}} \\
&= \frac{(\alpha + n + 1)^{\underline{h}} \cdot (\alpha + n + 1 - h)}
{h + 1}
+ \frac{(\alpha + n + 1)^{\underline{h}} \cdot (h + 1)}
{h + 1}
- \frac{\alpha^{\underline{h + 1}}}{h + 1} \\
&= \frac{(\alpha + n + 1)^{\underline{h}} \cdot (\alpha + n + 2)}
{h + 1}
- \frac{\alpha^{\underline{h + 1}}}{h + 1} \\
&= \frac{(\alpha + n + 2)^{\underline{h + 1}}}{h + 1}
- \frac{\alpha^{\underline{h + 1}}}{h + 1}
\end{align}$
This is exactly as claimed.
For your second sum, note that:
$$
(\alpha + k)^{\underline{h}} x^{\alpha + k - h}
= \frac{\mathrm{d}^h}{\mathrm{d} x^h} x^{\alpha + k}
$$
Thus the sum is essentially:
$\begin{align}
\sum\limits_{k=0}^n (\alpha + k)^{\underline{h}} r^k
&= r^{h - \alpha} \sum\limits_{k=0}^n (\alpha + k)^{\underline{h}} r^{\alpha + k - h} \\
&= r^{h - \alpha}
\left.
\frac{\mathrm{d}^h}{\mathrm{d} x^h} \sum\limits_{k=0}^n x^{\alpha + k}
\right|_{x = r} \\
&= r^{h - \alpha}
\left.
\frac{\mathrm{d}^h}{\mathrm{d} x^h} \left(x^\alpha \sum\limits_{k=0}^n x^k\right)
\right|_{x = r} \\
&=\left.r^{h-\alpha} \frac{d^h}{d x^h} \left( \frac{x^\alpha - x^{\alpha+n+1}}{1- x} \right)\right|_{x=r} \\
&=r^{h-\alpha} \sum\limits_{l=0}^h \binom{h}{l} \left(\alpha^{\underline{h-l}} r^{\alpha-(h-l)} - (\alpha+n+1)^{\underline{h-l}} r^{\alpha+n+1-(h-l)}\right) \cdot \frac{l!}{(1-r)^{l+1}} \\
&= \frac{1}{(1-r)^{h+1}} \sum\limits_{l=0}^h \binom{h}{l} \left(\alpha^{\underline{h-l}} r^{l} - (\alpha+n+1)^{\underline{h-l}} r^{n+1+l}\right) \cdot l! (1-r)^{h-l} \\
&= \frac{h!}{(1-r)^{h+1}} \sum\limits_{l=0}^h \frac{1}{(h-l)!} \left(\alpha^{\underline{h-l}} r^{l} - (\alpha+n+1)^{\underline{h-l}} r^{n+1+l}\right) \cdot (1-r)^{h-l} \\
&= \frac{h!}{(1-r)^{h+1}} \sum\limits_{l=0}^h \left(\binom{\alpha}{h-l} r^{l} - \binom{\alpha+n+1}{h-l} r^{n+1+l}\right) \cdot (1-r)^{h-l} \\
&= \frac{h!}{(1-r)} \left\{
\binom{\alpha}{h} \cdot
F_{2,1} \left[\begin{array}{ll}1 & -h \\ \alpha-h+1 \end{array};\frac{r}{r-1}\right]
- r^{n+1} \binom{\alpha+n+1}{h} \cdot
F_{2,1} \left[\begin{array}{ll}1 & -h \\ \alpha+n-h+2 \end{array};\frac{r}{r-1}\right]
\right\}
\end{align}$
This is quite ugly. The remaining sum is geometric, and can be expressed as a fraction. Leibnitz' formula for multiple derivatives of a product reduce that somewhat, but it is still a mess. If $h$ is small integer, perhaps a CAS gives somewhat manageable.
Note: It is a matter of taste whether an expression is ugly or not. In my opinion the final result has a closed form if we use hypergeometric functions.
Best Answer
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.