http://www.math.toronto.edu/barbeau/putnamalg.pdf
Let $n$ be a positive integer, and let $f_n(z) = n + (n-1)z + (n-2)z^2 + ... + z^{n-1}$. Prove that $f_n$ has no roots in the unit disk.
I know of the usual proof which involves defining $g(z) = (z - 1)f(z)$, but I wanted to try it in a different way.
Note that $f_n(z) = n + z(f_{n-1}(z))$.
Then suppose $f_n(z) = 0$ for some $z$ with $|z| \le 1$.
We have $f_{n-1}(z) = \frac{-n}{z}$. We can repeat this to obtain expressions for $f_{n-2}(z), f_{n-3}(z), ... f_1(z)$. For example, we get
$f_{n-2}(z) = \frac{-n - (n-1)(z)}{z^2}$
$f_{n-3}(z) = \frac{-n - (n-1)z - (n-2)z^2}{z^3}$
Eventually, we will get $f_1(z) = \frac{-n - (n-1)z - (n-2)z^2 - ... - z^{n-2}}{z^{n-1}}$
If $|z| \le 1$, this is impossible, as it implies
$|f_1(z)| = |\frac{n + (n-1)z + ... + z^{n-2}}{z^{n-1}}| \ge (*) \space n + (n-1) + ... + 1 > 1$
Which is impossible as $f_1(z) = 1$.
The step (*) uses the triangle inequality and the facts that when $|z| \le 1$, $\frac{|a|}{|z|} >= |a|$ and $|z^2| \le |z|$.
Is this correct? Is there any way I can make a similar argument in a less messy way?