Find the function of integer numbers $\sum_{n=0}^{\infty }\frac{n^k}{n!}=f(k) \cdot e$ Find the function of integer numbers

$$\sum_{n=0}^{\infty }\frac{n^k}{n!}={f(k)}\cdot e$$

I took many values of $k$ and I found the following results
$$\sum_{n=0}^{\infty }\frac{n^1}{n!}=e$$
$$\sum_{n=0}^{\infty }\frac{n^2}{n!}=2e$$
$$\sum_{n=0}^{\infty }\frac{n^3}{n!}=5e$$
$$\sum_{n=0}^{\infty }\frac{n^4}{n!}=15e$$
$$\sum_{n=0}^{\infty }\frac{n^5}{n!}=52e$$
and so on
I think these numerical values are right, so I tried to find the function $f(k)$.
Can help me to find the function $f(k)$ and then prove the above series
 A: Define the following sequence 
$$h_{k+1}(x) = xh_k'(x)$$
with $h_0(x) = e^x$. Then (by induction)
$$h_k(x) = \sum_{n=0}^\infty \frac{n^kx^{n}}{n!}$$
and from this it follows (again by induction) that $$h_k(1) \equiv \sum_{n=0}^\infty \frac{n^k}{n!} = f(k) e$$ where $f(k)$ are integers. To find an expression for $f(k)$ define
$$g(x,t) = \sum_{n=0}^\infty \frac{h_{n}(x)t^n}{n!}$$
then
$$g(x,t) = \sum_{n=0}^\infty\sum_{k=0}^\infty \frac{(kt)^n}{n!k!}x^k = e^{xe^t}$$
and by taking $x=1$ it follows that
$$\sum_{n=0}^\infty \frac{f(n)x^n}{n!} = e^{e^x -1}$$
which is the exponential generation function for the Bell numbers.
A: The function $F(x) = e^x$ is given by
$$F(x) = \sum_{n=0}^\infty \frac{x^n}{n!}$$
and thus has $k$th derivative
$$F^{(k)}(x) = \sum_{n=0}^\infty n(n - 1) \cdots (n - (k-1)) \frac{x^n}{n!} = k!\sum_{n=0}^\infty \binom{n}{k} \frac{x^n}{n!}.$$
(I'm being very cavalier here about the sum, but standard uniform convergence arguments will make the approach rigorous.) But $F^{(k)}(x) = F(x)$, so
$$\sum_{n=0}^\infty \binom{n}{k} \frac{x^n}{n!} = \frac{1}{k!} f^{(k)}(1) = \frac{e}{k!}.$$
Rewriting the polynomials $n^k$ in terms of the $\binom{n}{m}$ gives the required result.
