Show there exists a value such that each partial sum equals its limit in modulus 
For each $n \in \mathbb{N}_0$, and for all $z \in \mathbb{C}$, define $$p_n(z) := \sum^{n}_{k=0} {z^k \over {k!}}.$$
Show that for all $r > 0$ and for all $n \in \mathbb{N}_0$, there exists $z \in \mathbb{C}$ with $|z|=r$ such that $|p_n(z)| = |e^z|$.

There is a first part to this problem that asks one to show that for all $r > 0$, there exists $N \in \mathbb{N}$ such that for all $n > N$, $p_n$ has no zeros in $B(0,r)$.  This follows pretty immediately from $e^z$ being the limit of $p_n$ and Hurwitz's Theorem.  So far, I haven't found this useful.
I'm also curious as to what the assignment $z(n;r)$ looks like, and if it's a function.  I know $z(0;r) = ir$ thus far, and these also appear to be the only solutions for $n=0$.
Edit: tag for Bessel functions added since one of the answers makes an interesting use of them for a more technical approach to this problem.
 A: We can apply the maximum modulus principle to the function $f_n(z) = p_n(z)\cdot e^{-z}$. We have $f_n(0) = 1$, and $f_n$ is not constant, hence there is a $z_r$ with $\lvert z_r\rvert = r$ and $\lvert f_n(z_r)\rvert > 1$. Furthermore, we have $0 < p_n(r) < e^r$ and hence $0 < f_n(r) < 1$.
$f_n$ attains values with modulus less than $1$ as well as values with modulus greater than $1$ on the circle $\lvert z\rvert = r$, hence by continuity also values with modulus $1$. But such values are precisely the values with $\lvert p_n(z)\rvert = \lvert e^z\rvert$.
A: As stated in the comments, we have that $\|p_n(r)\|=p_n(r)<e^r=\|e^r\|$, hence we just need to prove that:
$$\exists\, z=r\,e^{i\theta}:\quad \|p_n(z)\|^2 > \|e^z\|^2 = e^{2r\cos\theta}\tag{1}$$
then apply a continuity argument. If we prove that:
$$\int_{0}^{\pi}\|p_n(r e^{i\theta})\|^2\sin\theta\,d\theta > \int_{0}^{\pi}e^{2r\cos\theta}\sin\theta\,d\theta=\frac{\sinh(2r)}{r}\tag{2}$$
then $(1)$ just follows. Notice that:
$$\|p_n(r e^{i\theta})\|^2 = p_n(re^{i\theta})\cdot p_n(r e^{-i\theta})=\sum_{j=0}^{n}\sum_{k=0}^{n}\frac{r^{j+k}}{j!\,k!}e^{(j-k)i\theta},\tag{3}$$
$$\int_{0}^{\pi}\cos(m\theta)\sin\theta\,d\theta = \left\{\begin{array}{rcl}0 &\text{if}& m\equiv 1\!\!\!\pmod{2}\\ -\frac{2}{m^2-1}&\text{if}&m\equiv 0\!\!\!\pmod{2}\end{array}\right.\tag{4}$$
so:
$$\begin{eqnarray*}\int_{0}^{\pi}\|p_n(r e^{i\theta})\|^2\sin\theta\,d\theta&=&2\sum_{j=0}^n\frac{r^{2j}}{j!^2}-\sum_{k=1}^{\lfloor n/2\rfloor}\frac{4}{4k^2-1}\sum_{j=0}^{n-2k}\frac{r^{2j+2k}}{j!(j+2k)!}\\&>&2\,I_0(2r)-\sum_{k=1}^{+\infty}\frac{4\,I_{2k}(2r)}{4k^2-1}\end{eqnarray*}\tag{5}$$
where $I_n(z)$ is the modified Bessel function of the first kind. 
By using the integral representation for such functions, from $(5)$ we have:
$$\begin{eqnarray*}\int_{0}^{\pi}\|p_n(r e^{i\theta})\|^2\sin\theta\,d\theta&>&2\,I_0(2r)-\frac{1}{\pi}\int_{0}^{\pi}e^{2r\cos \theta}\sum_{k=1}^{+\infty}\frac{4\cos(2k\theta)}{4k^2-1}\,d\theta\\&=&2\,I_0(2r)-\frac{1}{\pi}\int_{0}^{\pi}e^{2r\cos \theta}\left(2-\pi\sin\theta\right)d\theta\\&=&\int_{0}^{\pi}e^{2r\cos\theta}\sin\theta\,d\theta=\frac{\sinh(2r)}{r},\end{eqnarray*}$$
as wanted.

Maybe it is possible to prove $(1)$ by choosing a different weight-function $\psi:(0,\pi)\to\mathbb{R}^+$ and proving the weigthed $L^2((0,\pi))$ inequality:

$$\int_{0}^{\pi}\|p_n(r e^{i\theta})\|^2\psi(\theta)\,d\theta>\int_{0}^{\pi}e^{2r\cos\theta}\psi(\theta)\,d\theta\tag{2*}$$

$\psi(\theta)=\sin(\theta)$ was just the most natural choice for me, since I numerically checked that $(2*)$ does not hold with $\psi(\theta)=1$.

This problem also arose another question on Meta.
