TITLE: Sums of sums of sums of...of numbers
QUESTION [6 upvotes]: If we introduce the following notation
$$S_r^q=\overbrace{\sum_{a_{r-1}=1}^q\sum_{a_{r-2}=1}^{a_{r-1}}\cdots\sum_{a_1=1}^{a_2}\sum^{a_1}}^{\mbox{a total of $r$ sums}}1$$
for example, $S^q_1=q$, $S^q_2=q(q+1)/2$ and so on, then one can show that 
$$S^p_{q-1}=S^q_{p-1},$$
where $p$ and $q$ are positive integers. What is the simplest proof of this? I know of one but suspect that there exists simpler ones. Is there any generalisation of this statement. Can somebody also direct me to some references on related material. Thanks a lot in advance!

REPLY [11 votes]: Using the formula
$$
\sum_{k=0}^n\binom{k}{j}=\binom{n+1}{j+1}\tag{1}
$$
we get inductively that
$$
S_q^p=\binom{p+q-1}{q}\tag{2}
$$
Thus, we have
$$
S_{q-1}^p=\binom{p+q-2}{q-1}
$$
and
$$
S_{p-1}^q=\binom{p+q-2}{p-1}
$$
and your identity is just the common symmetry identity $\displaystyle\binom{n}{k}=\binom{n}{n-k}$.

Inductive Reasoning

$(2)$ holds for $q=\color{#C00000}{1}$ because $\displaystyle S_{\color{#C00000}{1}}^p=p=\binom{p+\color{#C00000}{1}-1}{\color{#C00000}{1}}$
Assume $(2)$ holds for some $q$. By definition, assumption, and $(1)$,
$$
\begin{align}
S_{q+1}^p
&=\sum_{k=1}^pS_q^k\\
&=\sum_{k=1}^p\binom{k+q-1}{q}\\
&=\binom{p+q}{q+1}\\
&=\binom{p+(q+1)-1}{q+1}
\end{align}
$$
Therefore, $(2)$ holds for $q+1$.