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?

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 $\frac{(m+N+1)!}{(m+1)N!}$.
