TITLE: Show that $p_n(a)\neq 0$ if $|a|=n$
QUESTION [8 upvotes]: I am working the next problem:

Consider the polynomials 
  $$
p_n(z)=\sum_{j=0}^{n}\frac{z^j}{j!}
$$
  For $n \geq 2$, show that if $a \in \mathbb{C}$ is such that $|a|=1$ or $|a|=n$, then $p_n(a)\neq 0$

The case $|a|=1$ has already been answer here on the last comment of this question, however I still have no clue for the case $|a|=n$. Any help? I have tried a lot of things, as you can see on the other question but nothing good.

REPLY [4 votes]: This can be shown using a sharpened Kakeya-Enestrom theorem due to Anderson, Saff, and Varga.  The paper (unfortunately missing pages 12 and 13) is available from Saff's webpage: PDF link.
In particular, in the notation of that paper $\beta = n$, so by Corollary 2 all zeros of $p_n(z)$ satisfy $|z| < n$.
We note that the same result can be applied to the reciprocal polynomial $q_n(z) = z^n p_n(1/z)$.  In that case it says that all zeros of $q_n(z)$ lie in $|z|<1$ and hence all zeros of $p_n(z)$ lie in $|z|>1$.