Why is the (-1)-th coefficient of $f^n f'$ equal to 0, without dividing by $n+1$?

Solution 1:

I don't think it becomes that messy to do explicitly. The coefficient you want is given by $$\sum_{i_1+i_2+\dots+i_n+i_{n+1}=0}\left(i_{n+1}\prod_k f_{i_k}\right) \, ;$$ the $k$th index on this sum comes from the $k$th factor of $f$ in the original product, while the $(n+1)$st index comes from the derivative factor.

Now, combine all the terms that contain the same combination $\prod_k f_{i_k}$, permuted somehow. Each $i_k$ will show up as a coefficient exactly $n!$ times, so the condition $\sum i_k = 0$ means that each of these combined terms vanishes.

Solution 2:

If your ring is an integral domain, remember that the characteristic of the ring must be a prime number, and that in that case, by the binomial theorem applied inductively, $$ (a_1 + \dots + a_k)^p = a_1^p + a_2^p + \dots + a_k^p, $$ (this is known as the freshman's dream, which is a common mistake in rings of characterstic $0$ made by starters in algebra, but is true in characteristic $p$) which means that proving the theorem for $n = 0, \dots, p-1$ is sufficient, because for $n = p$ a prime number, we get \begin{align*} f(x)^p (f'(x)) & = \left(\sum_{n \in \mathbb Z} f_n x^n \right)^p \left( \sum_{m \in \mathbb Z} m f_m x^{m-1} \right) \\ & = \left(\sum_{n \in \mathbb Z} f_n^p x^{np} \right) \left( \sum_{m \in \mathbb Z} m f_m x^{m-1} \right) \\ & = \sum_{n, m \in \mathbb Z} f_n^p f_m m x^{np+m-1} \end{align*} The coefficients before $x^{-1}$ are those with an exponent of the form $np+m-1 = -1$, but in characteristic $p$, this means $m = 0$. You can readily see that the coefficients of all such terms ($f_n^p f_m m$) will indeed vanish.

I don't know if you get so easily past the $p^{th}$ powers ($n = 2p, 3p, \dots$), but this is an idea to get around it. In rings which are not integral domains I have no idea what to do.

Hope that helps,