The book by Herbert Wilf "generatingfunctionology" has the exercise:
Let A(x) be a power series with $A(0) = 1$, and let $B(x) = A(x)^k$. It is desired to compute the coefficients of $B(x)$, without raising $A(x)$ to any powers at all. Use the '$xD\log$' method to derive a recurrence formula that is satisfied by the coefficients of $B(x)$.
The answer then gives
$$nb_n = \sum_{j=1}^n(j(k+1)-n)a_jb_{n-j}\quad\text{for }n\geq 1,$$
with $b_0 =1$.
However, I don't see how to arrive at this solution. After the '$xD \log$' method one has $kA'B=AB'$, so
$$k\sum_na_nnx^{n-1}\sum_nb_nx^n = \sum_na_nx^n\sum_nb_nnx^{n-1}.$$
Then with the Cauchy-Product formula one gets
$$k\sum_n\sum_jja_jb_{n-j}x^{n-1} = \sum_n\sum_j(n-j)a_jb_{n-j}x^{n-1}.$$
But I fail to see how one continues from here? One idea includes to change $kA'B=AB'$ beforehand to $(1/ A)kA'B=B'$ by using the reciprocal of $A$?
Thans a lot!