I have a polynomial associate to a number: $$\begin{align} k&=1 &n-1\\ k&=2 &n^2-2n +2\\ k&=3 &n^3-3n^2+6n+6\\ k&=4 &n^4-4n^3+12n^2-24n+24 \end{align}$$
and in general$$ P(k+1)=n^k-(k+1)P(k)$$
Can those polynomials be expressed in terms of negative binomial coefficients?
In other words I am searching for a general formula to express those polynomials
On OEIS, there is an unsigned integer sequence with your coefficients.
Now they have one good mathematica routine there, namely
CoefficientList[Series[(1 + x)^m, {x, 0, 20}], x]* Table[n!, {n, 0, m}], {m, 0, 10}] // GridI made this nested table, which I prefer, straight from the formula on OEIS (top of the page) to do the same thing, and modified it to include the alternating sign behavior:
Table[Table[(-1)^k*n!/(n - k)! , {k, 0, n}], {n, 0, 8}] // GridThe result is
\begin{array}{ccccccccc} 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -2 & 2 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -3 & 6 & -6 & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -4 & 12 & -24 & 24 & \text{} & \text{} & \text{} & \text{} \\ 1 & -5 & 20 & -60 & 120 & -120 & \text{} & \text{} & \text{} \\ 1 & -6 & 30 & -120 & 360 & -720 & 720 & \text{} & \text{} \\ 1 & -7 & 42 & -210 & 840 & -2520 & 5040 & -5040 & \text{} \\ 1 & -8 & 56 & -336 & 1680 & -6720 & 20160 & -40320 & 40320 \\ \end{array}
I am taking your formula to be
$$P(k+1)=n^{k+1}-(k+1)P(k).$$
Thus I have that with the alternating sign behavior, the coefficients of the $n^{th}$ row $R_n$ of your sequence is given by the sequence formula
$$R_n=\mathop{\left\{ (-1)^k\frac{n!}{(n-k)!} \right\}}\limits_{k=0}^{n}.$$
For example, row 4 is
$$R_4=\mathop{\left\{ (-1)^k\frac{4!}{(4-k)!} \right\}}\limits_{k=0}^{4}=\{ 1,-4,12,-24,24 \}.$$
Changing variable names so that we can use your $n$ in the polynomial base along with the coefficients, I manage to exactly duplicate your polynomial terms, and say that the $t^{th}$ row polynomial is given by the sum
$$R_t=\sum_{k=0}^t(-1)^k\frac{t!}{(t-k)!}n^{t-k}.$$
For completeness, I conclude with another Mathematica routine to duplicate your rows:
Table[TraditionalForm[ Sum[(-1)^k*t!/(t - k)!*n^(t - k), {k, 0, t}]], {t, 1, 8}] // TableFormThe output of that is
\begin{array}{l} n-1 \\ n^2-2 n+2 \\ n^3-3 n^2+6 n-6 \\ n^4-4 n^3+12 n^2-24 n+24 \\ n^5-5 n^4+20 n^3-60 n^2+120 n-120 \\ n^6-6 n^5+30 n^4-120 n^3+360 n^2-720 n+720 \\ n^7-7 n^6+42 n^5-210 n^4+840 n^3-2520 n^2+5040 n-5040 \\ n^8-8 n^7+56 n^6-336 n^5+1680 n^4-6720 n^3+20160 n^2-40320 n+40320 \\ \end{array}