While performing a computation, I came across a problem in which I needed to apply the MacLaurin expansion of $e^2$
$$e^2=\sum_{n=0}^{\infty}\frac{2^n}{n!}$$
I then noticed that similar expansions generated the following results $$\sum_{n=0}^{\infty}\frac{2^nn}{n!}=2\sum_{n=0}^{\infty}\frac{2^n}{n!}=2e^2\tag{1}$$ $$\sum_{n=0}^{\infty}\frac{2^nn^2}{n!}=6\sum_{n=0}^{\infty}\frac{2^n}{n!}=6e^2\tag{2}$$ $$\sum_{n=0}^{\infty}\frac{2^nn^3}{n!}=22\sum_{n=0}^{\infty}\frac{2^n}{n!}=22e^2\tag{3}$$ $$\sum_{n=0}^{\infty}\frac{2^nn^4}{n!}=94\sum_{n=0}^{\infty}\frac{2^n}{n!}=94e^2\tag{4}$$
and that if I found the pattern for the coefficients of $2,6,22,$ and $94$ above, I would be able to find the value for some arbitrary $k$
$$\sum_{n=0}^{\infty}\frac{2^nn^k}{n!}=?\sum_{n=0}^{\infty}\frac{2^n}{n!}=?e^2\tag{5}$$
However, it isn't clear to me why the coefficients are $2,6,22,$ and $94$. I am therefore curious to understand two related questions
Why do the coefficients in $(1)$ through $(4)$ come out to be $2,6,22,$ and $94$? Is this an application of geometric series or something similar which I have forgotten?
Is there a deeper result that would generate the coefficients for any power of $n^k$? I would assume that this wouldn't be independent to $e^2$ and would work for any MacLaurin expansion of $e^x$ for any integer $x$.
The Touchard polynomials may be defined as $$T_k(x)=e^{-x}\sum_{n=0}^\infty\frac{x^nn^k}{n!}$$ and represent the $k$th moment of a Poisson-distributed random variable with mean $x$. The sequence $2,6,22,94$ you inquire about comes from fixing $x=2$: $$T_k(2)=e^{-2}\sum_{n=0}^\infty\frac{2^nn^k}{n!}$$ $$T_k(2)e^2=\sum_{n=0}^\infty\frac{2^nn^k}{n!}$$ But we also have $$T_k(x)=\sum_{n=0}^kS(k,n)x^n$$ so we have a finite summation as the formula for the coefficients: $$T_k(2)=\sum_{n=0}^kS(k,n)2^n$$ where $S(n,k)$ is the Stirling number of the second kind. The exponential generating function for $T_k(2)$ is $e^{2(e^x-1)}$.