I'm looking for a short and precise proof of the following identity;
$$\left(\sum_{k=0}^\infty \frac{C_k}{k!}x^k\right)^n=\sum_{k=0}^\infty\left[ \sum_{(j_1+...+j_n=k)}\binom{k}{j_1,...,j_n}\frac{C_{j_1}\cdot...\cdot C_{j_n}}{k!}\right]x^k$$
I've got the idea through brute force of smaller convolutions when $n=2,3,4$ and I've used Mathematica to show that the identity holds for the previous cases. But I was hoping for something compact.
The coefficient of $x^k$ in $$\left(\sum_{k\ge 0}\frac{C_k}{k!}x^k\right)^n$$ is the sum of all possible products of the form
$$\prod_{i=1}^n\frac{C_{j_i}}{j_i!}\;,$$
where the $j_i$ are non-negative integers whose sum is $k$. Thus, it’s
$$\begin{align*} \sum_{j_1+\ldots+j_n=k}\frac{C_{j_1}\cdot\ldots\cdot C_{j_n}}{j_1!\cdot\ldots\cdot j_n!}&=\sum_{j_1+\ldots+j_n=k}\left(\frac{k!}{j_1!\cdot\ldots\cdot j_n!}\cdot\frac{C_{j_1}\cdot\ldots\cdot C_{j_n}}{k!}\right)\\\\ &=\sum_{j_1+\ldots+j_n=k}\binom{k}{j_1,\ldots,j_n}\frac{C_{j_1}\cdot\ldots\cdot C_{j_n}}{k!}\;. \end{align*}$$