Prove a sum of sums equals n choose k In some research I'm doing, I've come across some coefficients I'm calling $\alpha^{n}_{j}$, where
$$ \alpha^{n}_{j} = \sum_{k_1 = 1}^{n} \sum_{k_2 = 1}^{n-k_1} ... \sum_{k_j = 1}^{n - k_1 - k_2 -... - k_{j-1}}1, \quad \text{ where   }\quad \sum_{m=1}^{p<1}1 = 0 $$
For every coefficient I have computed, I've found $\alpha_{j}^{n} = {n\choose j}$. However, I am struggling to prove this. Does anybody have any advice? 
By looking at some combinations, I've convinced myself of the following:
$$\alpha_{j}^{n} = 1 + \sum_{k = 1}^{n-j} \sum_{l = 1}^{k} {k-1 \choose l-1}{j \choose l}, \quad \text{ where   }\quad{p<m \choose m}=0$$
I'm not sure what the next step to proving this would be though, or if this is the right track.
 A: Your initial expression could be rewritten as
$$\sum_{k_1+k_2+\cdots+k_j\leq n}1$$
where additionally you require each $k_i\geq 1$, i.e. it counts the number of ordered tuples $(k_1,\ldots,k_j)$ with sum at most $n$.
Now there is a transformation between such tuples and subsets of $1,...,n$ of size $j$. The numbers $k_1,k_1+k_2,k_1+k_2+k_3,...$ are all different and in the range $1,...,n$. Conversely for any set of $j$ numbers between $1$ and $n$ there is such a tuple: make $k_1$ the smallest number in the list, $k_2$ the difference between the smallest and second-smallest, and so on. So there is a 1-to-1 correspondence between these tuples and combinations of $j$ numbers from $1,...,n$ - but the number of those is $\binom nj$ by definition.
A: Hint: 
$$
\eqalign{
  & \sum\limits_{0\, \le \,k\, < \,n} 1  = n = \left( \matrix{
  n \cr 
  1 \cr}  \right)  \cr 
  & \sum\limits_{0\, \le \,k\, < \,n} {\sum\limits_{0\, \le \,j\, < \,n - k} 1 }  = \sum\limits_{0\, \le \,k\, < \,n} {\left( \matrix{
  n - k \cr 
  1 \cr}  \right)}  = \sum\limits_{0\, < \,k\, \le \,n} {\left( \matrix{
  k \cr 
  1 \cr}  \right)}  = \sum\limits_k {\left( \matrix{
  n - k \cr 
  n - k \cr}  \right)\left( \matrix{
  k \cr 
  k - 1 \cr}  \right)}  = \left( \matrix{
  n + 1 \cr 
  n - 1 \cr}  \right) = \left( \matrix{
  n + 1 \cr 
  2 \cr}  \right)  \cr 
  & \quad  \vdots  \cr} 
$$
while, starting the sums from $1$
$$
\eqalign{
  & \sum\limits_{1\, \le \,k\, \le \,n} 1  = n = \left( \matrix{
  n \cr 
  1 \cr}  \right)  \cr 
  & \sum\limits_{1\, \le \,k\, \le \,n} {\sum\limits_{1\, \le \,j\, \le \,n - k} 1 }  = \sum\limits_{1\, \le \,k\, \le \,n} {\left( \matrix{
  n - k \cr 
  1 \cr}  \right)}  = \sum\limits_{0\, \le \,k\, \le \,n - 1} {\left( \matrix{
  k \cr 
  1 \cr}  \right)}  = \sum\limits_k {\left( \matrix{
  n - 1 - k \cr 
  n - 1 - k \cr}  \right)\left( \matrix{
  k \cr 
  k - 1 \cr}  \right)}  = \left( \matrix{
  n \cr 
  n - 2 \cr}  \right) = \left( \matrix{
  n \cr 
  2 \cr}  \right) \cr} 
$$
