Show that $a_nb_n$ is an integer The following question is taken from here exercise $3:$

For each $n\geq 1,$ let 
  $$a_n = \sum_{k=1}^\infty \frac{k^n}{k!}, \,\, b_n = \sum_{k=1}^\infty (-1)^n \frac{k^n}{k!}.$$
  Show that $a_nb_n$ is an integer. 

My attempt: 
Clearly $b_n = (-1)^n a_n.$
So for odd $n,$ we have $a_n+b_n = 0$ and for even $n,$ we have $a_n=b_n.$
So I thought I wanted to show that $(a_n)^2$ is an integer, but it turns out that it is not by Wolfram Alpha.
However, when I keyed in the required product for $n=1,$ it turns out that it is not an integer. Is the question correct?
EDITED: It seems that $(-1)^n$ should be $(-1)^k.$ In other words, the correct question is as follows: 

For each $n\geq 1,$ let 
  $$a_n = \sum_{k=1}^\infty \frac{k^n}{k!}, \,\, b_n = \sum_{k=1}^\infty (-1)^k \frac{k^n}{k!}.$$
  Show that $a_nb_n$ is an integer. 

EDITED AGAIN: I notice that 
$$a_n=\sum_{k=1}^\infty \frac{k}{k!} = e$$
$$a_n=\sum_{k=1}^\infty \frac{k^2}{k!} = 2e$$
$$a_n=\sum_{k=1}^\infty \frac{k^3}{k!} = 5e$$
$$a_n=\sum_{k=1}^\infty \frac{k^4}{k!} = 15e$$
and $b_n$ contains an integer of $e^{-1}.$ 
So $a_nb_n$ will cancel $e,$ left with integers only. 
So I wanted to prove a general formula for 
$$\sum_{k=1}^\infty \frac{k^n}{k!} = ?$$
using induction. However, I fail to obtain a general form. 
 A: Observe that for $n \geq 1$
$$A_n(x) := \sum_{k=1}^{\infty} \frac{k^n}{k!} x^k = \left(x\frac{d}{dx}\right)^n e^x = p_n(x) e^x$$
where $p_n(x)$ is a polynomial with integer coefficients.
Therefore $a_n = A_n(1)$ is an integer times $e$. In the same way $b_n$ is an integer times $e^{-1}$.
Thus $a_n b_n$ is an integer.

Elaborating according to a question in the comments.
Deriving the Maclauring expansion of $e^x$ we get
$$
\frac{d}{dx}e^x 
= \frac{d}{dx} \sum_{k=0}^{\infty} \frac{1}{k!}x^k
= \sum_{k=1}^{\infty} \frac{k}{k!}x^{k-1}
$$
(Notice the change of lower sum limit since the term for $k=0$ vanishes.)
Multiplying with $x$ then gives
$$
x\frac{d}{dx}e^x 
= \sum_{k=1}^{\infty} \frac{k}{k!}x^{k}
$$
Deriving once again gives
$$
\frac{d}{dx}x\frac{d}{dx}e^x 
= \sum_{k=1}^{\infty} \frac{k^2}{k!}x^{k-1}
$$
and after multiplication by $x$
$$
x\frac{d}{dx}x\frac{d}{dx}e^x 
= \sum_{k=1}^{\infty} \frac{k^2}{k!}x^{k}
$$
Generally we can write
$$
\left(x\frac{d}{dx}\right)^n e^x 
= \sum_{k=1}^{\infty} \frac{k^n}{k!}x^{k}
$$
Now lets look at the left hand sides.
For $n=0$ we have
$\left(x\frac{d}{dx}\right)^0 e^x = e^x = p_0(x) e^x$
where $p_0(x)=1$.
For $n=1$ we have
$$x\frac{d}{dx} e^x = x e^x = p_1(x) e^x$$
where $p_1(x)=x$.
For $n=2$ we have
$$\left(x\frac{d}{dx}\right) e^x = x \frac{d}{dx}(x e^x) = x(x+1) e^x = p_2(x) e^x$$
where $p_2(x)=x(x+1) = x^2+x$.
It's obvious, and can easily be shown using induction, that every $p_n(x)$ will have (non-negative) integer coefficients.
A: Hint: Observe that 
$$\begin{align*} 
\sum_{k=1}^{\infty} \frac{k^{\underline{i}}}{k!} &= e \\[1ex]
\sum_{k=1}^{\infty} (-1)^k \frac{k^{\underline{i}}}{k!} & = (-1)^i e^{-1}
\end{align*} $$
for $i \geqslant 1$, where $k^{\underline{i}} = k \cdot (k-1) \cdot \ldots \cdot (k-i+1)$. We can always write 
$$k^n = \sum_{i=0}^n \alpha_i \cdot k^{\underline{i}}$$
where each $\alpha_i$ is an integer. This allows us to compute both sums 
$$\sum_{k=1}^{\infty} \frac{k^n}{k!}, \qquad \sum_{k=1}^{\infty} (-1)^k \frac{k^n}{k!}$$
and multiply them, and the result turns out to be an integer. 
