On the arithmetic of powers of subseries of the exponential series Let $p$ be a prime number and $q=p-1$. I’m trying to prove that the nonzero coefficients $a_{qk}$ ($k\ge1$) of the power series
$$ \sum_{k\ge1} a_{qk} z^{qk} := \left( \sum_{k\ge0} \frac{z^{qk+1}}{(qk+1)!} \right)^q $$
satisfy the congruence
$$ a_{qk} \cdot (qk)! \equiv -1 \mod p. $$
I’ve managed to work out that: There is a combinatorial formula for this power series. First of all, the series being raised to the $q$th power can be written
$$ f(z) = \frac1q \sum_{1\le i\le q} \zeta^{-i} e^{\zeta^iz} = \sum_{k\ge0} \frac{z^{qk+1}}{(qk+1)!} $$
where $\zeta$ is a primitive $q$th root of unity. And its $q$th power can be expressed in terms of the derivative
$$ f'(z) = \frac1q \sum_{1\le i\le q} e^{\zeta^iz} = \sum_{k\ge0} \frac{z^{qk}}{(qk)!} $$
by the formula
$$
    f(z)^q  = \frac1{q^{q}} \sum {q\choose i_1,\dots,i_{q}} {\big|C_q\cdot(i_1,\dots,i_q)\big|} \zeta^{-\big(1\cdot i_1+2\cdot i_2+\cdots+q\cdot i_{q}\big)} f'\Big(\big(i_1\zeta^1+\cdots+i_{q}\zeta^{q}\big)z\Big) $$
where the sum ranges over $C_{q}$-orbits of weak compositions $i_1+\cdots+i_{q}=q$, $i_k\ge0$, and where $\big|C_q\cdot(i_1,\dots,i_q)\big|$ denotes the size of the $C_q$-orbit of the weak composition.
It is straightforward to show that Fermat's little theorem extends to $\mathbf{F}_p[\zeta]/(\zeta^q-1)$ in the sense that the Frobenius is the identity, and it follows that $q$th powers in this ring are fixed by exponentiation by any positive integer power.
This reduces proving the claim to verifying it for the coefficient of $z^q$, i.e. verifying the congruence
$$
  -1 \equiv \sum {q\choose i_1,\dots,i_{q}} {\big|C_q\cdot(i_1,\dots,i_q)\big|} \cdot \zeta^{-\big(1\cdot i_1+2\cdot i_2+\cdots+q\cdot i_{q}\big)} \Big(i_1\zeta^1+\cdots+i_{q}\zeta^{q}\Big)^q \mod p.
$$
I’ve done this by computer for the primes up to $p=17$ but haven’t found a general argument.
 A: Your argument is in fact almost complete. It reduces the problem to checking the normalized coefficient $a_q$ of $z^q$ is congruent to $-1$ mod $p$, but since $$ f = z + \frac{z^{q+1}}{(q+1)!} + \dots ,$$ we have $$f^q = z^q + \frac{ q z^{2q}}{ (q+1)!} + \dots$$ and so $$a_q = q! \equiv -1 \mod p.$$

Here is an alternate "bijective" presentation of the same argument. We can interpret $a_n$ combinatorially as the number of $q$-colorings of the numbers from $1$ to $n$ such that each color is used a number of times congruent to $1$ mod $q$.
For $n \geq p$, there is an action of $\mathbb Z/p$ on this set of colorings by rotating the colorings of the numbers $n+1-p, n+2-p, \dots, n$. The number of colorings is congruent mod $p$ to the number of fixed points of this action. A coloring is fixed if the last $p$ numbers have the same color.
Since $q=p-1$, removing the last $q$ numbers in such a coloring doesn't affect the congruent mod $q$ of the number of times each coloring is used, and this gives a bijection between fixed points and colorings of the first $n-q$ numbers satisfying the same congruence condition.
So $a_n \equiv a_{n-q} \mod p$ for $n>q$ and $a_q$ is the number of $q$-colorings of $\{1,\dots, q\}$ with all colors occurring exactly once, which is $q! \equiv -1 \mod p$.
