**Proof.**
Proof of (1). If $\gamma $ is a divided power structure, then condition (2) (applied to $1$ and $n-1$ instead of $n$ and $m$) implies that $n \gamma _ n(x) = \gamma _1(x)\gamma _{n - 1}(x)$. Hence by induction and condition (1) we get $n! \gamma _ n(x) = x^ n$.

Assume $A$ is torsion free as a $\mathbf{Z}$-module. Proof of (2). This is clear from (1).

Proof of (3). Assume that $n! \gamma _ n(x) = x^ n$ for all $x \in I$ and $n \geq 1$. Since $A \subset A \otimes _{\mathbf{Z}} \mathbf{Q}$ it suffices to prove the axioms (1) – (5) of Definition 23.2.1 in case $A$ is a $\mathbf{Q}$-algebra. In this case $\gamma _ n(x) = x^ n/n!$ and it is straightforward to verify (1) – (5); for example, (4) corresponds to the binomial formula

\[ (x + y)^ n = \sum _{i = 0, \ldots , n} \frac{n!}{i!(n - i)!} x^ iy^{n - i} \]

We encourage the reader to do the verifications to make sure that we have the coefficients correct.

Proof of (4). Assume we have generators $x_ i$ of $I$ as an ideal such that $x_ i^ n \in (n!)I$ for all $n \geq 1$. We claim that for all $x \in I$ we have $x^ n \in (n!)I$. If the claim holds then we can set $\gamma _ n(x) = x^ n/n!$ which is a divided power structure by (3). To prove the claim we note that it holds for $x = ax_ i$. Hence we see that the claim holds for a set of generators of $I$ as an abelian group. By induction on the length of an expression in terms of these, it suffices to prove the claim for $x + y$ if it holds for $x$ and $y$. This follows immediately from the binomial theorem.
$\square$

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