General formula for iterated cumulative sum

The numbers in sequence $S_k$ are the binomial coefficients $\binom{m}{k}$; the $n$-th term of $S_k$ is $\binom{n-1+k}{k} = \frac{(n-1+k)!}{(n-1)!k!}$. One can prove this by using that $$ \binom{i}{i} + \binom{i+1}{i} + \cdots + \binom{i+j}{i} = \binom{i+j+1}{i+1} $$ for any $i,j \geq 0$.

For $k=1$ we find $P_1(n) = \binom{n-1+1}{1} = \binom{n}{1} = n$, for $k=2$ we find $P_2(n) = \binom{n-1+2}{2} = \binom{n+1}{2} = \frac{n(n+1)}{2}$, for $k=3$ we find $P_3(n) = \binom{n-1+3}{3} = \binom{n+2}{3} = \frac{n(n+1)(n+2)}{6}$, etcetera.

We can generalize your problem to arbitrary initial sequences.

Let $a_n\,(n=0,1,2,3,...)$ be a sequence of numbers. Define its iterated partial sums using the recurrence $$S^{(0)}_n=a_n,\quad S^{(k+1)}_n=\sum_{m=0}^n S^{(k)}_m,\tag1$$ so that we have, for example, $$\small\begin{array} &S^{(0)}_0=\color{green}{a_0}, &S^{(0)}_1=\color{blue}{a_1}, &S^{(0)}_2=\color{maroon}{a_2},&...\\ S^{(1)}_0=\color{green}{a_0}, &S^{(1)}_1=\color{green}{a_0}+\color{blue}{a_1}, &S^{(1)}_2=\color{green}{a_0}+\color{blue}{a_1}+\color{maroon}{a_2},&...\\ S^{(2)}_0=\color{green}{a_0}, &S^{(2)}_1=\color{green}{a_0}+(\color{green}{a_0}+\color{blue}{a_1}), &S^{(2)}_2=\color{green}{a_0}+(\color{green}{a_0}+\color{blue}{a_1})+(\color{green}{a_0}+\color{blue}{a_1}+\color{maroon}{a_2}),&...\\ S^{(3)}_0=\color{green}{a_0}, &S^{(3)}_1=\color{green}{a_0}+(\color{green}{a_0}+(\color{green}{a_0}+\color{blue}{a_1})),&... \end{array}\tag2$$ Now we can prove by induction that the following formula holds: $$S^{(k+1)}_n=\sum_{m=0}^n\binom{m+k}k\,a_{n-m}.\tag3$$