Just to be clear, what is sought is to find the general term
$a_n$ of the sequence
$$
\eqalign{
a_0&=e^{-m}, \cr
a_1&=e^{-m} (2-m), \cr
a_2&=\frac{1}{2} (m-4) m+1, \cr
a_3&=-\frac{1}{6} m ((m-6) m+6), \cr
a_4&=\frac{1}{24} m^2 ((m-4) m+12), \cr
a_5&=-\frac{1}{120} m^3 ((m-5) m+20), \cr
}
$$
Right?
There's not much of a pattern there. A factor of
$(-1)^n/n!$ Another factor of
$m^{n-2}$ for
$n\ge2$, which mysteriously becomes
$e^{-m}$ for
$n=0,1$, perhaps. Finally, again for
$n\ge2$, a factor of the form
$(m-b_n)m+n(n-1)$, where
$b_n$
is a seemingly random sequence,
$b_2=4$,
$b_3=6$,
$b_4=4$,
$b_5=5$.
Except for the factor
$e^{-m}$, the term
$a_1$ could be seen to fit the pattern of
$a_n$ for
$n\ge2$.
Then
$b_n=2n$ for
$n=1,2,3$ and
$b_n=n$ for
$n=4,5$ —
who knows what happens for
$n>5$?
Here's the sequence based on the pattern for
$a_2$,
$a_3$, if two terms constitute a pattern:
Table[
Subscript[a, n] == (-1)^n/n! m^(n - 2) ((m - 2 n) m + n (n - 1)),
{n, 0, 5}] // Column // TeXForm
$$
\eqalign{
a_0&=1, \cr
a_1&=2-m, \cr
a_2&=\frac{1}{2} ((m-4) m+2), \cr
a_3&=-\frac{1}{6} m ((m-6) m+6), \cr
a_4&=\frac{1}{24} m^2 ((m-8) m+12), \cr
a_5&=-\frac{1}{120} m^3 ((m-10) m+20), \cr
}
$$
In fact, we see that both
$a_0$ and
$a_1$ in the OP's sequence would fit this formula if the mysterious factor
$e^{-m}$ were a mistake and could be dropped from each term.
Where this sequence differs from the OP's suggests that the OP's sequence is much more complicated, that is, assuming it contains no errors. In my experience, I have to give FindSequenceFunction[] a couple dozen or more terms for it to discover a formula for a complicated sequence.
