Summation of Series with Factorials It is given that:
$$v_n = n(n+1)(n+2)\;...\;(n+m)$$
$$and$$
$$u_n = (n+1)(n+2)\;...\;(n+m)$$

$i.$ Verify that:
$$v_{n+1} - v_n = (m+1)(n+1)(n+2)\;...\;(n+m)$$

I started off by inspecting $v_n$:
$$v_1 = 1(2)(3)\;...\;(1+m) = (m+1)!$$
$$v_2 = 2(3)(4)\;...\;(2+m) = \frac{(m+2)!}{1}$$
$$v_3 = 3(4)(5)\;...\;(3+m) = \frac{(m+3)!}{2!}$$
$$\implies v_n = \frac{(m+n)!}{(n-1)!}$$
$$\implies v_{n+1} - v_n = \frac{(m+n+1)!}{n!} - \frac{(m+n)!}{(n-1)!}$$
That's as far as I could get. How can I verify the above equation?

$ii.$ Find $\sum_{n=1}^N u_n$ in terms of $m$ and $N$

Since $u_n$ can be written as $\frac{v_n}{n}$:
$$\implies \sum_{n=1}^N u_n = \sum_{n=1}^N \frac{v_n}{n} = \sum_{n=1}^N \frac{(m+n)!}{n(n-1)!} = \sum_{n=1}^N \frac{(m+n)!}{n!}$$
How can I sum the above equation given the factorial complication?
 A: For $i$, factor out $(n+1)\cdots(n+m)$ from $v_{n+1}-v_n$.
For $ii$, $\sum_{n=0}^Nu_n=\sum_{n=0}^N\frac{v_{n+1}-v_n}{m+1}$, using the results of $i$. Pull out the $m+1$ and sum with the telescoping series method, yielding $\frac{1}{m+1}(v_{N+1}-v_0)$, which equals $\frac{(m+N+1)!}{(m+1)N!}$.
A: As alex.jordan has aleady pointed out, there’s no need to work with factorials:
$$\begin{align*}
v_{n+1}-v_n&=\prod_{k=0}^m(n+1+k)-\prod_{k=0}^m(n+k)\\\\
&\overset{(1)}=\prod_{k=1}^{m+1}(n+k)-\prod_{k=0}^m(n+k)\\\\
&\overset{(2)}=(n+m+1)\prod_{k=1}^m(n+k)-n\prod_{k=1}^m(n+k)\\\\
&\overset{(3)}=\Big((n+m+1)-n\Big)\prod_{k=1}^m(n+k)\\\\
&=(m+1)\prod_{k=1}^m(n+k)\;,
\end{align*}$$
as desired. And $\prod_{k=1}^m(n+k)=u_n$, so we’ve established that $u_n=\frac1{m+1}(v_{n+1}-v_n)$. Thus,
$$\begin{align*}
\sum_{n=1}^Nu_n&=\frac1{m+1}\sum_{n=1}^N(v_{n+1}-v_n)\\\\
&=\frac1{m+1}\left(\sum_{n=1}^Nv_{n+1}-\sum_{n=1}^Nv_n\right)\\\\
&\overset{(1)}=\frac1{m+1}\left(\sum_{n=2}^{N+1}v_n-\sum_{n=1}^Nv_n\right)\\\\
&\overset{(2)}=\frac1{m+1}\left(v_{N+1}+\sum_{n=2}^Nv_n-\sum_{n=2}^Nv_n-v_1\right)\\\\
&\overset{(3)}=\frac1{m+1}(v_{N+1}-v_1)\;.
\end{align*}$$
In the steps marked $(1)$ I’ve shifted the index of the first product (or sum) by $1$ to give the general expressions inside the two products (or sums) identical forms. In the steps marked $(2)$ I’ve separated the extra factor (or term) from each of the two products (or sums), so that in what’s left the index runs over the same range of values. And in the steps marked $(3)$ I’ve combined the identical products (or sums). This is a fairly common pattern of manipulation.
It’s a good idea to learn perform manipulations like these on products and summations, but of course you can always write them out longhand with ellipses for the missing terms, as alex.jordan did in his first comment (with the blue terms).
