Roots of a polynomial 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 $p_n(a)\neq 0$

For the case $|a|=1$, I think i have a partial solution: Suppose that $a\in \mathbb{C}$ is such that $p_n(a)=0$ and $|a|=1$, then by the revers triangle inequality
$$
0 = | p_n(a)| > \left| 1 - \left|a+\frac{a^2}{2}+\cdots+\frac{a^n}{n!}\right|\right|
$$
which is a contradiction, because since $n\geq 2$, the RHS of the last inequality must be positive (I said partial solution because I am not sure how to prove this, I am almost sure that this follows because $| a + \cdots a^n/n!|>1$ if $n\geq 2$ but I can't prove that either) EDIT: According to the comments this is wrong, so my question now extends to both cases!. 
For the case $|a|=n$, i tried something similar but it gets worst.
My questions are: 1) Is my approach for the first case correct? If it is, how can I prove the details I am missing, and if is not how can I approach it? 2) How can I approach the second case ? Any help or hints will be very appreciated 
 A: Bringing down the solution in the comments given by User Kevin Arlin in order to take this question off the unanswered list.

Note that the given polynomial $p_n(z)$ is the $n$th order Taylor polynomial of $f(z) = e^z$. Suppose $\lvert a \rvert = 1$. On the one hand, we have $1/e \leq \lvert e^a \rvert \leq e$. On the other hand,
$$
\lvert e^a - p_n(a) \rvert = \left\lvert \sum_{m = n+1}^\infty \frac{a^m}{m!} \right\rvert \leq \sum_{m = n+1}^\infty \frac{1}{m!}.
$$
So, if $a$ were a zero of $p_n(z)$, then we would have
$$
\frac{1}{e} \leq \lvert e^a \rvert = \lvert e^a - p_n(a) \rvert \leq \sum_{m = n+1}^\infty \frac{1}{m!}.
$$
But, for $n \geq 2$, we have
$$
\sum_{m = n+1}^\infty \frac{1}{m!} \leq e - \frac{1}{0!} - \frac{1}{1!} - \frac{1}{2!} = e - 5/2.
$$
However,
$$
\frac{1}{e} > e - \frac{5}{2},
$$
a contradiction.
Hence, if $a$ is a zero of $p_n(z)$ for $n \geq 2$, then $\lvert a \rvert \neq 1$.
(It might be worth noting that at $n = 1$, we have $p_1(z) = 1 + z$, which does have a zero of unit modulus, namely $a = -1$. When $n = 0$, we have $p_0(z) = 1$, which has no zeros at all, but this is a trivial scenario.)

The case $\lvert a \rvert = n$ is more complicated, and is dealt with in this question: Show that $p_n(a)\neq 0$ if $|a|=n$.
