Let $$g(z) = \sum_{k=0}^n z^k = \frac{z^{n+1}-1}{z-1}$$
It is easy to see the roots of $g(z)$ lie on the unit circle $|z|= 1 $ and all of them are simple.
By Gauss-Lucas theorem, the roots of $g'(z)$ belong to the convex hull of the roots of $g$. Since the closed unit disk $|z| \le 1$ is convex, this convex hull is a subset of the closed unit disk.
Notice the convex hull is a $n$-gon which intersect the unit circle $|z| = 1$ only at the roots of $g(z)$. Since the roots of $g(z)$ are simple, none of them can be root of $g'(z)$. As a result, the roots of $g'(z)$ belongs to the open unit disk $|z| < 1$.
Since $zg'(z)$ differs from $g'(z)$ by only a root at $0$, the roots of
$$zg'(z) = \sum_{k=1}^n k z^k$$
belong to the open unit disk $|z| < 1$.
Apply Gauss-Lucas theorem again and then add a root at $z = 0$, we find the roots of
$$\left(z\frac{d}{dz}\right)^2 g(z) = \sum_{k=1}^n k^2 z^k$$
belong to the open unit disk $|z| < 1$.
Repeat this process one more time, we find all the zeros of
$$\left(z\frac{d}{dz}\right)^3 g(z) = \sum_{k=1}^n k^3 z^k = P_n(z)$$
lie inside the unit circle.
Notes
There are other ways to arrive at same conclusion. In particular, we can use following results:
Let $\displaystyle\;f(z) = \sum_{k=0}^m a_k z^k$ be any polynomial with real and positive coefficients.
If the coefficients $a_k$ are non-descending,
$$0 < a_0 \le a_2 \le \cdots \le a_m$$
then roots of $f(z)$ belong to the closed unit disk $|z| \le 1$.
If the coefficients $a_k$ are non-ascending,
$$ a_0 \ge a_1 \ge \cdots \ge a_m > 0$$
then roots of $f(z)$ lie outside the open unit disk (i.e. $|z| \ge 1$ for all the roots)
In general, the roots of $f(z)$ belong to the closed annulus
$$\min_{1\le k \le m} \frac{a_{k-1}}{a_k} \le |z| \le \max_{1 \le k \le m}\frac{a_{k-1}}{a_k}$$
Since it is easy to derive any one of these results from the other two,
these results are typically treated as a single theorem known as the Eneström-Kakeya Theorem.
For a proof of the first result, see answers of a related question. In particular, the answer by Ayman Hourieh which uses Rouché's theorem.
Back to the problem at hand. It is easy to see we can rewrite $P_n(z)$ as $z f(z)$ for some polynomial $f(z)$ with real and positive coefficients. Apply the third result, we immediately find aside from a root at $z = 0$, the remaining $n-1$ roots of $P_n(z)$ lies within the closed annulus $$\frac18 \le |z| \le \left(\frac{n-1
}{n}\right)^3 < 1$$
