For the common exercise here are two similar proofs, but one is completely elementary:
Proof 1
This is the elementary proof.
Consider
$\frac{p(z)(z-1)}{z^{n+1}} = a_n + \sum_{k=0}^{n-1} \frac{a_k - a_{k+1}}{z^{n-k}} - \frac{a_0}{z^{n+1}} = a_n - \sum_{k=0}^{n-1} \frac{a_{k+1} - a_k}{z^{n-k}} - \frac{a_0}{z^{n+1}} $
Thus $\displaystyle \left|\frac{p(z)(z-1)}{z^{n+1}}\right| \ge a_n - \sum_{k=0}^{n-1} \frac{a_{k+1} - a_k}{|z^{n-k}|} - \frac{a_0}{|z^{n+1}|}$
using $\displaystyle |z_1 - z_2| \ge |z_1| - |z_2|$
Now if $\displaystyle |z| \gt 1$, then $\displaystyle -\frac{a_{k+1} - a_k}{|z^{n-k}|} \gt -(a_{k+1} - a_k)$ and $\displaystyle -\frac{a_0}{|z^{n+1}|} \gt - a_0$
Thus
$\displaystyle \left|\frac{p(z)(z-1)}{z^{n+1}}\right| \gt a_n - \sum_{k=0}^{n-1} (a_{k+1} - a_k) - a_0 = 0$
Thus if $\displaystyle |z| \gt 1$, then $\displaystyle |p(z)| \neq 0$ and so all the roots of $\displaystyle p(z)$ satisfy $\displaystyle |z| \le 1$.
Proof 2
Similar to the above consider the polynomial
$g(z) = p(z)(1-z) + a_n z^{n+1} = a_0 + \sum_{k=0}^{n-1} (a_{k+1} - a_k) z^{k+1}$
Now given any $\displaystyle c > 0$, we have that, for any $\displaystyle z$ on the boundary of the unit circle,
$|g(z)| \le a_0 + \sum_{k=0}^{n-1} (a_{k+1} - a_k) = a_n \lt |(a_n + c)z^{n+1}|$
Thus if $\displaystyle f(z) = (a_n + c) z^{n+1}$, we have that $\displaystyle |g(z)| \lt |f(z)|$ on the boundary of the unit disk and thus both $\displaystyle g(z)-f(z)$ and $\displaystyle f(z)$ have the same number of roots satisfying $\displaystyle |z| \lt 1$, by using Rouche's Theorem.
Thus for any $\displaystyle c > 0$, the polynomial $\displaystyle p(z)(1-z) - c z^{n+1}$ has all its roots within the unit circle, i.e. the root $\displaystyle w$ with maximum modulus satisfies $\displaystyle |w| \lt 1$.
Now it is a well known fact that the eigenvalues of a complex matrix are continuous and thus the function of the root with the maximum modulus is continuous (though I am not completely sure of this). Taking limits as $\displaystyle c \to 0$ we get the result that the roots of $\displaystyle p(z)(1-z)$ all satisfy $\displaystyle |z| \le 1$.