This question concerns Herstein's proof (in Topics in Algebra) that $e$ is transcendental. If you have the book, it's Theorem 5.2.1, page 218, or there's a transcription here: https://sites.math.washington.edu/~palmieri/Courses/2005/Math403/transcendental.pdf.
In the final step, he goes from $$\epsilon_i = \frac{-ie^{i\left(1-\theta_i\right)}\left(i\theta_i\right)^{p-1}\left(1-i\theta_i\right)^p \dotsm \left(n-i\theta_i\right)^p}{(p-1)!}$$ to $$\lvert\epsilon_i\rvert \leq \frac{e^n n^p \left(n!\right)^p}{(p-1)!}$$ I can see where most of this estimate comes from, but am having difficulty with the $\left(n!\right)^p$ part. This seems to assume that $$\left\lvert\left(1-i\theta_i\right) \dotsm \left(n-i\theta_i\right)\right\rvert \leq n! \qquad (*)$$ but some of the terms inside the absolute value will be negative (the $\theta_i$ are between 0 and 1, but $i$ can range from $1$ to $n$), so we can't just multiply the inequalities $1-i\theta_i < 1$, $2-i\theta_i < 2$, etc. So can anyone clarify how the inequality $(*)$ arises? Thanks very much in advance for assistance.
In Herstein's book it reads $(1-i\theta_i) \cdots (n-i \theta_i)$, not $(1-i\theta_1) \cdots (n-i \theta_i)$. I think this is easily seen to be bounded by $n!$.
In fact, let $k$ be the smallest integer with $1 \leq k \leq n$ and $k - i \theta_i > 0$. Then $(n-i \theta_i) \cdots (k - i \theta_i) \leq (n-k+1)!$. For the remaining factors we have $\left| (1-i\theta_i) \cdots (k-1 - i\theta_i)\right| \leq (k-1)!$, where we say the empty product equals $1$. The total product therefore is bounded by $(n-k+1)!(k-1)!$ which equals $n!/\binom{n}{k-1}\leq n!$.