This is an exercise in Atiyah-MacDonald (Exercise 5.2). I have shown that if $f \in A[[x]]$ ($A$ commutative with unity) is nilpotent, then $f$'s coefficients are all nilpotent. The next question in the exercise is: "Is the converse true?"
I think not, but it would be wonderful if someone else could check my solution. I'm aware that there are solutions elsewhere, but I'm trying to avoid looking at those.
Let $A = \mathbf{Z}[z_2,z_3,\dots]/I$, where $I = (z_i^2 : 2 \leq i < \infty)$. Further define $z_0 = z_1 = 0$; clearly every element of $\{z_i\}_{i=0}^\infty$ is nilpotent. Now define $f(x) = \sum_{i=0}^\infty z_ix^i$.
Let $N = \sum_{i=3}^{k-2} i = {k-1 \choose 2} - 3$. Then after expanding, the coefficient of $x^N$ in $f^k$ is:
$${ N \choose 1, 1,\dots,1} z_3z_4\dots z_{q-2} + \sum(\cdots)$$
where $\sum(\cdots)$ is the remaining terms. Since ${N \choose 1,\dots,1} \neq 0$ and $z_3z_4\dots z_{q-2} \neq 0$, it follows that the coefficient of $x^N$ in $f^k$ is nonzero, hence $f^k \neq 0$. This construction holds for every $k \in \mathbf{N}$, so $f$ is not nilpotent.