The divided polynomial algebra over a field Let $\Gamma_R[\alpha]$ denote the divided polynomial algebra over $R$; that is, the quotient of the free $R$-algebra $R\langle \alpha_1,\alpha_2,\cdots \rangle$ by the relations $$\alpha_n \cdot \alpha_m = \binom{n+m}{n}\alpha_{n+m},$$ with $t_0=1$. I am particularly interested in the case where $R = \mathbb{F}_p$. 
The claim is that 
$$\Gamma_{\mathbb{F}_p}[\alpha] \simeq \bigotimes_{k \ge 0} \mathbb{F}_p[\alpha_{p^i}]/(\alpha_{p^i}^p)$$
(Note: I presume the tensor product is over $\mathbb{F}_p$ here?)
The proof is given is Hatcher's algebraic topology book pp. 286-287. I can understand what he is doing (sort-of), but I can't see how it all combines to give a proof.
First he claims that $\Gamma_{\mathbb{F}_p}[\alpha] = \Gamma_{\mathbb{Z}}[\alpha]\otimes \mathbb{F}_p$. I don't see why this is the case?
Then he claims that this is equivalent to the statement

$\ast$ The element $\alpha_1^{n_0}\alpha_{p}^{n_1} \cdots \alpha_{p^k}^{n_k}$ in $\Gamma_\mathbb{Z}[\alpha]$ is divisible by $p$ iff $n_i \ge p$ for some $i$. 

which I don't see follows from the above. Now, as he states, we can use the product relation above to get 
$$\alpha_1^{n_0}\alpha_{p}^{n_1} \cdots \alpha_{p^k}^{n_k} = m \alpha_n$$ for $n = n_0+n_1p+\cdots n_k p^k$ and some integer $m$, and then the question is if $p$ divides $m$. 
His other fact is 

$\ast \ast \alpha_n \alpha_{p_k}$ is divisible by $p$ iff $n_k=p-1$, assuming $n_i < p$ for each $i$. 

I am OK with this - this follows easily from Lucas' theorem.
How does $\ast \ast$ imply $\ast$? It is meant to be via an inductive argument  by multiplying on the right by $\alpha_{p^i}$, but I can't quite see exactly what I should be proving.
Any tips appreciated!
 A: To see that $\ast \ast$ implies $\ast$:
First, it should be clear that $\alpha_i^p$ is divisible by $p$ (in $\Gamma_\mathbb{Z}[\alpha]$), since, for some integer $c$ $$\alpha_i^p = c \cdot \alpha_{pi}$$ and one of the factors in $c$ is $\binom{p i}{(p-1) i}$ which is divisible by $p$ (the top has more factors of $p$ than the bottom, so this follows from Lucas' theorem). This proves one direction of $\ast$.  
For the other direction, we must show that if all $n_i < p$, then the indicated product is not divisible by $p$.  We proceed by induction on $k$.  For the base case, note that $\alpha_1^n = c \alpha_n$, and the constant $c$ is the product of binomial coefficients all of whose top entries are $\le n < p$, hence $c$ is not divisible by $p$.  For the inductive step, the product under consideration can be written (by the inductive hypothesis) as $$c \cdot a_m \cdot a_{p^k}^{n_k}$$ where $m = n_0 + n_1 p + \ldots n_{k-1}p^{k-1}$, and $c$ is not divisible by $p$.  But by $\ast \ast$, $a_m a_{p^k}^{n_k}$ is not divisible by $p$ because the coefficient of $p^k$ in $m$ is 0, not $p-1$.  Hence the entire product is not divisible by $p$. 
