If $p_0(n),p_1(n)\ldots$ is a sequence of polynomials satisfying $$\sum_{k \geq 0}p_k(n) \frac {x^k}{k!}= \left ( \sum_{ k \geq 0}p_k(1) \frac {x^k}{k!} \right )^n$$ then $$p_k(m+n)= \sum_{i=0}^{k} {k\choose i}p_i(m)p_{k-i}(n), \; k \geq 0$$
Any help would be appreciated.
Hint : Try to write $$\sum_{k\geq 0}p_{k}(n+m)\frac{x^{k}}{k!}$$ as the product of 2 exponential generating functions and use the Cauchy product (binomial convolution) of 2 exponential generating functions.