Proper divisor of order of $\mathbb{F}^*_q$ must divide element its in prime decomposition? I'm puzzled by a step in a proof in Lidl & Niederreiter's Finite Fields.  For the record, it's the proof of Theorem 2.8, "For every finite field $\mathbb{F}_q$ the multiplicative group $\mathbb{F}^*_q$ of nonzero elements of $\mathbb{F}_q$ is cyclic."  I'm not looking for a proof of the theorem per se; there are other proofs of it that I understand.  I want to understand this one.
Let $b_i = a_i^{h/p_i^{r_i}}$, where $a_i$ is some element of $\mathbb{F}_q$ that is not a root of $x^{h/p_i}-1$, and $p_i^{r_i}$ is an element in the prime factor decomposition of the order of $\mathbb{F}^*_q$, which is $h=p_1^{r_1}p_2^{r_2}\cdots p_m^{r_m}$.
Lidl & Niederreiter write:

We claim that element $b=b_1 b_2 \cdots b_m$ has order $h$.  Suppose, on the contrary, that the order of $b$ is a proper divisor of $h$, and is therefore a divisor of at least one of the $m$ integers $h/p_i$ ....

I understand everything in the proof up until this point, and everything after it.  What I don't understand is why, if the order of $b$ were a proper divisor of $h$, it must be a divisor of one of the elements $p_i^{r_i}$.  Why couldn't the order of $b$ be, say, $p_1^2 p_3$, if for example $h=p_1^3 p_2 p_3^2$ ?  After all, $\mathbb{F}^*_q$ is not a field, and a subgroup wouldn't necessarily be a field, so there's no reason on that score that the order of $b$ subgroup must be a prime power.
I've been puzzling over this now for a couple of weeks.  I suspect that there's something obvious that I am just not seeing--some dots I haven't connected--or that I am just confused about something.  I've gone back a few pages through theorems, and gone back to the groups section of this book and other books.  I've looked at different proofs of the same theorem, but they don't include the $b_1 b_2 \cdots b_m$ construction--at least not explicitly, and I don't see how it's implicit.  I thought that the answer might lie in theorems about orders of cyclic subgroups [L&N 1.15(ii), 1.15(iii)], and that may still be the case, but I don't see why.
(I'd be happy to be pointed to a dupe question.  The same idea may appear in another question, but I have not found it yet.)
 A: At the suggestion of commenters, I'm answering my own question, with thanks to @WhatsUp.
As @WhatsUp noted in a comment, Lidl & Niederreiter did not suggest that if the order of $b$ is a proper divisor of $h$, it must be a divisor of one of the prime power factors $p_i^{r_i}$ of $h$.  What they said was that the order of $b$ would have to be a divisor of $h/p_i$ for some $i$.
On the assumption that ord($b$) is a proper divisor of $h$, considering each instance of $p_i$ in $p_i^{r_i}$ as a distinct factor of $h$, at least one such $p_i$ instance is not a factor of ord($b$): that is, ord($b$) is a product of fewer instances of primes than $h$.  Thus since $h = p_1^{r_1} p_2^{r_2} \cdots p_m^{r_m}$,
$$h/p_i = p_1^{r_1} p_2^{r_2} \cdots p_i^{r_i - 1} \cdots p_{m-1}^{r_{m-1}} p_m^{r_m}$$
is such a quantity, created by dividing out $p_i$ from $h$. The expansion of $h/p_i$ shows that Lidl & Niederreiter did not exclude the possibility that ord($b$) could be a divisor of multiple factors $p_j^t, p_k^u$, etc. of $h$; it might divide any combination of factors in $p_1^{r_1} p_2^{r_2} \cdots p_i^{r_i - 1} \cdots p_{m-1}^{r_{m-1}} p_m^{r_m}$, were it a proper divisor of $h$.
