General form for the series expansion of $e$ I've found a lot of series expansions of the Napier's constant. I was wondering if a general form for this could be devised. They all have a trend and similarities. I've been trying but I've been screwing up as usual.


*

*$\Large{\sum\limits_{n=0}^{\infty} \frac{1}{(n - k)!} = e}$

*$\Large{\sum\limits_{n=1}^{\infty} \frac{1}{(n - k)!} = e - 1}$
where $k$ belongs to the set of natural numbers.


*

*$\Large{\sum\limits_{n = 2}^{\infty} \frac{1}{n!} = e - 2}$

*$\Large{\sum\limits_{n = 0}^{\infty} \frac{1}{(n + 1)!} = e - 1}$

*$\Large{\sum\limits_{n = 1}^{\infty} \frac{1}{(n + 1)!} = \sum\limits_{n = 0}^{\infty} \frac{1}{(n + 2)!} =  e - 2}$

*$\Large{\sum\limits_{n = 0}^{\infty} \frac{n}{n!} = e}$

*$\Large{\sum\limits_{n = 1}^{\infty} \frac{n^2}{n!} = \sum\limits_{n = 1}^{\infty} \frac{n^2}{n!} =  2e }$

*$\Large{\sum\limits_{n = 1}^{\infty} \frac{n^3}{n!} = \sum\limits_{n = 1}^{\infty} \frac{n^3}{n!} =  5e }$

*$\Large{\sum\limits_{n = 1}^{\infty} \frac{n^4}{n!} = \sum\limits_{n = 1}^{\infty} \frac{n^4}{n!} =  15e }$
I was wondering if you guys could help me out and give me a general equation that encapsulates all of this. A brief description of how it works might also be of interest to me.
Thank you, in advance, for your efforts.
 A: 
$$e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$$

example 7): $$\sum_{n=1}^{\infty}\frac{n^{2}}{n!}=\sum_{n=0}^{\infty}\frac{n+1}{n!}=\sum_{n=0}^{\infty}\frac{n}{n!}+e=2e$$
where 6) is used
A: You didn't specify what $k$ is in the first two descriptions but the correct forms would be the ones with $k=0$. All your formulas can be derived from the first:
$$
e=\sum_{n=0}^\infty \frac{1}{n!}=1+1+\frac{1}{2}+\ldots.
$$
Number 2 omits the first summand in the series so the result is $e-1$. Number 3 omits the first and the second. Number 4 is just an index shift away from number 2. I'll do one of the others:
$$
\sum_{n=1}^\infty \frac{n^2}{n!}=\sum_{n=1}^\infty \frac{n}{(n-1)!}=\sum_{n=0}^\infty\frac{n+1}{n!}=\sum_{n=0}^\infty\frac{n}{n!}+\sum_{n=0}^\infty\frac{1}{n!}=e+e.
$$
A: the formula $4,5,6,7$ can be proved the following way.
$$\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x$$
differentiate with respect to $x$
 $$\sum_{n=0}^{\infty}\frac{nx^{n-1}}{n!}=e^x$$
multiply both side by $x$
 $$\sum_{n=0}^{\infty}\frac{nx^{n}}{n!}=xe^x$$
repeat this cycle of differentiating with respect to $x$ and multiplying with $x$ $3$ more times to get.
 $$\sum_{n=0}^{\infty}\frac{n^4x^{n}}{n!}=xe^x(x^3+6x^2+7x+1)$$
now substitute $x=1$ to get
$$\sum_{n=0}^{\infty}\frac{n^4}{n!}=15e$$
