Multinomial Expansion of Finite Series to a Power. I am having some trouble understanding the multinomial theorem, and I wonder if someone could confirm if my solution is correct. Ultimately, I want to write out the solution in an expanded form in order to improve my understanding.
Using multinomial theorem we can rewrite $\left[\sum_{i=0}^{m-1}z^{i}a_{i}\right]^{q}$ as,

$$\left[\sum_{i=0}^{m-1}z^{i}a_{i}\right]^{q} =\sum_{i_{1}+i_{2}+\cdots+i_{m-1}=q}\dbinom{q}{i_{1},i_{2},\ldots,i_{m-1}}\prod_{t=1}^{m-1}z^{i_{t}}a_{i_{t}},$$

where $\dbinom{q}{i_{1},i_{2},\ldots,i_{m-1}}=\dbinom{q}{i_{1}!i_{2}!\cdots i_{m-1}!},$

$$=\sum_{i_{1}=0}^{m-1}\sum_{i_{2}=0}^{m-1}\cdots\sum_{i_{q}=0}^{m-1}\dbinom{q}{i_{1}!i_{2}!\cdots i_{m-1}!}\prod_{t=1}^{m-1}z^{i_{t}}a_{i_{t}}.$$

 A: No. Some issues. One, you're indexing your summands from $0$ to $m-1$, so the lower limit of your products should be $0$ not $1$. Two, you substituted into the multinomial theorem incorrectly. Three, the formula for the multinomial coefficient is
$$\binom{q}{i_0,\cdots,i_{m-1}}=\frac{q!}{i_0!\cdots i_{m-1}!}\color{Red}{\ne \binom{q}{i_0!\cdots i_{m-1}!}}. $$
The binomial coefficient you've written (I made it red above) is incorrect. Notice $i_0!\cdots i_{m-1}!$ is almost surely bigger than $q$ anyway, so that binomial coefficient is usually $0$.
Fourth, in your second display formula you've written $\sum_{i_1}\cdots\sum_{i_{m-1}}$ but you can't just drop the condition that the sum the $i$ terms must be $q$.
The multinomial theorem says
$$ \left(\sum_{t=0}^{m-1} c_t\right)^q=\sum_{i_0+\cdots+i_{m-1}=q}\binom{q}{i_0,\cdots,i_{m-1}}\prod_{t=0}^{m-1}c_t^{i_{\large t}} $$
If we substitute $c_t=a_tz^t$ then this becomes
$$ \sum_{i_0+\cdots+i_{m-1}=q}\binom{q}{i_0,\cdots,i_{m-1}}\prod_{t=0}^{m-1} \left(a_tz^t\right)^{i_{\large t}}.$$
