Show that the zeros of $\sum_{k=1}^n kz^{k-1}$ are inside the unit disc This is a homework problem that I wasn't able to solve, and I feel a little silly about it (the assignment has been submitted). The problem:

Let $P(z) = 1 + 2z + 3z^2 + \cdots + nz^{n-1}$. By considering $(1-z)P(z)$, show that zeros of $P(z)$ are contained inside the unit disc.

This is from Chapter 1 of Bak's Complex Analysis. A solution should only use algebraic manipulations of complex numbers (in other words, this problem is very easy to solve by employing the Gauss-Lucas theorem, but that's not "on the table" -- complex differentiation hasn't been defined in the scope of what's permissible).
What I've done:
$$\begin{align*}
(1-z)P(z) &= 1+z+z^2+\cdots+z^{n-1}-nz^n \\
 &= \frac{z^n-1}{z-1}-nz^n.
\end{align*}$$
Let $z$ be a zero of $P(z)$; then it is a zero of $(1-z)P(z)$. Consequently,
$$\begin{align*}
nz^n &= \frac{z^n-1}{z-1} \\
nz^n(z-1) &= z^n-1 \\
nz^n(z-1)+1 &= z^n
\end{align*}
$$
Then
$$|z^n| = |z|^n = |nz^n(z-1)+1|.$$
From the triangle inequality,
$$
\begin{align*}
|z|^n &\le |nz^n(z-1)| + |1| \\
|z|^n &\le n|z^n||z-1| + |1| \\
n|z|^n(|z-1|) &\le 1 \\
|z|^n &\le \frac{1}{n|z-1|}
\end{align*}
$$
At this point, I am stuck. I feel like I went down the wrong path, or maybe I just made an arithmetical error, but I can't find where.

Now you take the contradiction that $|z|> 1$, and as n is increasing, you will get a contradiction and so $|z|\le1$
