Homework 2 collaboration area

Here's some interesting stuff:

$ \sum_{n=1}^N 1 = \dfrac11N $

$ \sum_{n=1}^N n = \dfrac12N\left(N+1\right) $

$ \sum_{n=1}^N n\left(n+1\right) = \dfrac13N\left(N+1\right)\left(N+2\right) $

       $ \vdots $                  $ \vdots $

From the observation, we can assume the following formula is true:

$ \sum_{n=1}^N \dfrac{\left(n+k\right)!}{\left(n-1\right)!} = \dfrac1{k+2}\cdot\dfrac{\left(N+k+1\right)!}{\left(N-1\right)!}\quad \mathrm{for}\;k\in\mathbb{N} $