Prove that $\sum_{q=0}^{d-r}\sum_{s=r+q}^{d}{{\binom {r-1+q}{r-1}}(r-1)!s}=\sum_{s=r}^{d}{\binom{s}{r}(r-1)!s}$ by sum manipulation [closed]

The first trick is to switch the order of summation, which is often a very fruitful trick in simplifying double combinatorial sums. To do this, we first use an Iverson bracket to write the double sum in a way so the inner sum bounds do not depend on the outer sum variable. The expression $[s\ge r+q]$ is $1$ if the condition $s\ge r+q$ is true, $0$ otherwise. In the third equation, we get rid of the Iverson bracket by modifying the new inner summation, essentially the first trick in reverse.

$$ \begin{align} \sum_{q=0}^{d-r}\sum_{s=r+q}^{d}{{\binom {r-1+q}{r-1}}(r-1)!s} &=\sum_{q=0}^{d-r}\sum_{s=r}^{d}{{\binom {r-1+q}{r-1}}(r-1)!s}[s\ge r+q]\\ &\hspace{1cm}\searrow\hspace{-.5cm}\swarrow\\ &=\sum_{s=r}^{d}\sum_{q=0}^{d-r}{{\binom {r-1+q}{r-1}}(r-1)!s}[s\ge r+q]\\ &=\sum_{s=r}^{d}\sum_{q=0}^{\color{blue}{s-r}}{{\binom {r-1+q}{r-1}}(r-1)!s}\\ &=\sum_{s=r}^{d}(r-1)!s\sum_{q=0}^{s-r}{{\binom {r-1+q}{r-1}}} \end{align} $$ Finally, the inner summation $\sum_{q=0}^{s-r}{{\binom {r-1+q}{r-1}}}$ simplifies to $\binom{s}r$ using the hockey stick identity.