I'm currently trying to understand the proof of e being transcendental in Julian Havil The Irrationals, but not sure how the following is concluded.
A polynomial of degree $mp+p-1$ is defined on the interval $0<x<m$ by $$ f(x) = \frac{x^{p-1}(x-1)^{p}(x-2)^p\dots(x-m)^p}{(p-1)!} $$
and then a function \begin{equation} F(x) = \sum_{k = 0}^{mp+p-1}f^{(k)}(x) \end{equation}
the sum of the derivatives essentially, stopping at the degree as everything after would be $0$. Next he looks to evaluate $F(x)$ in two cases, $F(r)$, $r>0$, and $F(0)$ where $r$ is the positive integers up to $m$.
For the first case he states that the only way for the $k$th derivative to be non zero is for $k=p$, in which case theres a factor of $p!$ in the numerator of the derivative which cancels the $(p-1)!$ to leave $p$ in the numerator. This means that $F(r)$ for $r>0$ is an integer and has $p$ as a factor.
I see why the derivative must be $0$ for all $k<p$, and also see why it wont be for $k=p$. But for $p<k<pm+p-1$ why is it $0$? The factors have been differentiated out and we should be left with a polynomial?
Similarly, for the second case. He considers the $k$th derivative evaluated at $0$. Stating this is only non zero if $k=p-1$. But why is it zero if $p-1<k<pm+p-1$?