I wanted to know why is this equality true for all $k$ : $$\sum_{r \leq k} \sum_{j_1 + \dots + j_r = k} \frac{1}{r!j_1\dots j_r} = 1$$ and was told to look at the $z^k$ term in the power series expansion of $$\dfrac{1}{1 - z}= \exp(- \ln(1- z))= \exp \big( \sum_{j \geq 1} \frac{z^j}{j}\big) = \sum_{r\geq 0} \frac{\big( \sum_{j \geq 1} \frac{z^j}{j}\big)^r }{r!}$$
I thought of taking $z=0$ but its not the right way, can someone see how is the equality true ?
If $|z|<1$, then \begin{align*}\sum_{k=0}^\infty z^k &=\frac{1}{1 - z}= \exp(- \log(1- z))= \exp\! \bigg( \sum_{j = 1}^\infty \frac{z^j}{j}\bigg) \\ &= \sum_{r= 0}^\infty \frac{\Big( \sum_{j = 1}^\infty \frac{z^j}{j}\Big)^r }{r!}=\sum_{k=0}^\infty \left(\sum_{r \leq k} \sum_{j_1 + \ldots + j_r = k} \frac{1}{r!j_1\cdots j_r} \right) z^k.\end{align*} Equating the coefficients of like powers of $z$ yields the desired identity.