We can considerably simplify this multiple sum since we obtain for $n\geq 1$:
\begin{align*}\color{blue}{\sum_{m_1=0}^{n-1}}&\color{blue}{\sum_{m_2=0}^{m_1-1}\dots\sum_{m_{n-2}=0}^{m_{n-3}-1}\sum_{m_{n-1}=0}^{m_{n-2}-1}m_{n-1}}\\&=\sum_{m_1=0}^{n-1}\sum_{m_2=0}^{m_1-1}\dots\sum_{m_{n-2}=0}^{m_{n-3}-1}\sum_{m_{n-1}=0}^{m_{n-2}-1}\sum_{m_{n}=0}^{m_{n-1}-1}1\tag{1}\\&=\sum_{0\leq m_{n}<m_{n-1}<\cdots<m_2<m_1\leq n-1}1\tag{2}\\&\,\,\color{blue}{=1}\tag{3}\end{align*}
Comment:
In (1) we write $m_{n-1}$ as sum: $m_{n-1} =\sum_{m_n=0}^{m_{n-1}-1}1$.
In (2) we use another typical notation and write the index range as inequality chain.
In (3) we observe the index range contains exactly one $n$-tuple: $(0,1,2,\ldots,n-2,n-1)$.
