What is this polynomial series? I've come across this simple, yet peculiar series of polynomials wich I can't quite find a general formula for.
It goes as such:
$$1$$
$$x+1$$
$$x^2+x+2$$
$$x^3+x^2+2x+6$$
$$x^4+x^3+2x^2+6x+24$$
$$\vdots$$
I'm finding these polynomials by looking at a term in the $n^{th}$ integral of Ei(x).
For example, the 5th integral of Ei(x) is;
$$\frac{1}{120}(x^5Ei(x)-e^x(x^4+x^3+2x^2+6x+24))$$
 A: Let $F_0(x) = \text{Ei}(x)$.
It seems $F_n(x) = \text{Ei}(x) \frac{x^n}{n!} - \frac{e^x}{n!} P_n(x)$ where $P_n$ is a polynomial of degree $n-1$ for $n \ge 1$ and $F_n'(x) = F_{n-1}(x)$,
which translates to 
$$ P_n'(x) + P_n(x) = P_{n-1}(x) + x^{n-1}$$ 
I get
$$
\eqalign{P_1(x) &= 1\cr
P_2(x) &= x+1\cr
P_3(x) &= {x}^{2}+x+2\cr
P_4(x) &= {x}^{3}+{x}^{2}+2\,x+6\cr
P_5(x) &= {x}^{4}+{x}^{3}+2\,{x}^{2}+6\,x+24\cr
P_6(x) &= {x}^{5}+{x}^{4}+2\,{x}^{3}+6\,{x}^{2}+24\,x+120\cr
P_7(x) &= {x}^{6}+{x}^{5}+2\,{x}^{4}+6\,{x}^{3}+24\,{x}^{2}+120\,x+720\cr}$$
It seems $$P_n(x) = \sum_{m=0}^{n-1} (n-1-m)! x^m$$
which should be easy to prove by induction.
A: From the given polynomials, my guess is,
$$p_n(x)=\sum_{i=0}^n (n-i)!\cdot x^i$$
with the list starting from $n=0$
A: Now that the question has been edited, my old answer does not hold, but this new sequence has an easier pattern:
Every diagonal is:
1, mult by 1 , mult by 2, mult by 3, mult by 4,... 
so, $1$, $1 \times 1 = 1$, $1 \times 2 = 2$, $2 \times 3 = 6$, $6 \times 4 = 24$ ...
So it is just $(n-1)!$ as the nth diagonal entry on any given diagonal.
A: Those are factorials : $1=0!,1=1!,2=2!,6=3!,24=4!,...$

A: Old answer before question was edited: 
The difference in the earlier question was that the last coefficient of the last polynomial was $4$ and not $24$ and the last coefficient of the second-to-last polynomial was 2 and not 6. 
This is not a complete answer but an interesting pattern: One interesting observation is that any given coefficient in a row is the sum of all coefficients of a previous row whose row number is also smaller than or equal to the smallest prime divisor of the row number. (The converse also holds - the sum of coefficients in a smaller row whose row number is less than or equal to the smallest prime dividing a row will appear as a coefficient)
For example, in row $2$, all coefficients are $1$ since the only coefficient in row $1$ was $1$. In row $3$, we have a coefficient $2$ since the sum of coefficients in row $2$ was $2$, and we have a coefficient $1$ since the sum of coefficients in row $1$ was $1$. In row $4$, the smallest prime divisor is $2$, and so we are allowed only coefficients $1$ and $2$. In row $5$, we have coefficients $6$ (from row 4), $4$ (from row 3), $2$ (from row 2), and $1$ (from row 1). 

A bit of fun: This question and my answer reminded me a bit of this comic: http://spikedmath.com/comics/062-the-iq-test.png :) 
