Recurrence formula for sums I know that $$\sum_{n_{k-1}=1}^{n_k} \sum_{n_{k-2}=1}^{n_{k-1}} ... \sum_{n_0=1}^{n_1} 1=\frac{1}{k!}\prod_{i=0}^{k-1} (n_k+i)=\frac{(n_k+k-1)!}{k!(n_k-1)!}=\binom{n_k+k-1}{k}$$
But what if I have:
1/ $\sum_{n_{k-1}=1}^{n_k} \sum_{n_{k-2}=1}^{n_{k-1}} ... \sum_{n_0=1}^{n_1} n_0$ ?
2/ $\sum_{n_{k-1}=1}^{n_k} \sum_{n_{k-2}=1}^{n_{k-1}} ... \sum_{n_0=1}^{n_1} n_0(n_0+1)$ ?
Is there any formula to simplify it?
 A: OP's identity
\begin{align*}
A_k(N)=\sum_{n_{k-1}=1}^{N} \sum_{n_{k-2}=1}^{n_{k-1}} \cdots\sum_{n_0=1}^{n_1} 1=\binom{N+k-1}{k}\tag{1}
\end{align*}
depends on the upper limit  $N (\geq 1)$ of the outermost sum and the number $k$ of the involved $\Sigma$-symbols  indicated by the naming of the bounded variables $n_0,\ldots,n_{k-1}$. In fact the  naming of the bounded indices is not essential, as the right hand side of (1) shows.
We can calculate both expressions OP is asking for by using (1).

Since
  \begin{align*}
\sum_{n_0=1}^{n_1}1=n_1
\end{align*}
  we obtain from (1) for $N\geq 1$
  \begin{align*}
\sum_{n_{k}=1}^{N}\sum_{n_{k-1}=1}^{n_{k}} \cdots\sum_{n_1=1}^{n_2} \color{blue}{n_1}
&=\sum_{n_{k}=1}^{N} \sum_{n_{k-1}=1}^{n_{k}} \cdots\sum_{n_1=1}^{n_2} \color{blue}{\sum_{n_0=1}^{n_1}1}\\
&=A_{k+1}(N)\\
&=\binom{N+k}{k+1}
\end{align*}

$$ $$ 

Since
  \begin{align*}
\sum_{n_1=1}^{n_2}\sum_{n_0=1}^{n_1}1=\sum_{n_1=1}^{n_2}n_1=\frac{n_2(n_2+1)}{2}
\end{align*}
  we obtain from (1) for $N\geq 1$
  \begin{align*}
\sum_{n_{k+1}=1}^{N}&\sum_{n_{k}=1}^{n_{k+1}} \cdots\sum_{n_2=1}^{n_3} \color{blue}{\frac{n_2(n_2+1)}{2}}\\
&=\sum_{n_{k+1}=1}^{N} \sum_{n_{k-1}=1}^{n_{k}} \cdots\sum_{n_2=1}^{n_3} \color{blue}{\sum_{n_1=1}^{n_2}\sum_{n_0=1}^{n_1}1}\\
&=A_{k+2}(N)\\
&=\binom{N+k+1}{k+2}
\end{align*}

