Proper divisor of order of $\mathbb{F}^*_q$ must divide element its in prime decomposition?

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 $h$.