Show that $\sum\limits_{k=1}^\infty\frac{(-1)^{k+1}}{k}x^{k-1}$ converges uniformly on $[0,1]$

Show that $\sum\limits_{k=1}^\infty\frac{(-1)^{k+1}}{k}x^{k-1}$ converges uniformly on the closed interval $[0,1]$.

It is clear that it converges pointwisely and we already know that a power series converges uniformly on each closed interval which is a subset of its radius of convergence, here $(-1,1)$. However, the closed interval $[0,1]$ is not a subset of $(-1,1)$. So it is not clear if the series of functions converges uniformly.

My apporach:

Let be $f_n:[0,1]\to\mathbb{R}$ with $f_n(x):=\sum\limits_{k=1}^n\frac{(-1)^{k+1}}{k}x^{k-1}$, then we know that it is differentiable and $f_n'(x):=\sum\limits_{k=1}^n\frac{(-1)^{k+1}}{k(k-1)}x^{k-2}$. We see that\begin{align*}&\sum\limits_{k=1}^n\left|\frac{(-1)^{k+1}}{k(k-1)}x^{k-2}\right|\leq\sum\limits_{k=1}^n\frac{x^{k-2}}{k(k-1)}\leq\sum\limits_{k=1}^n\frac{1}{k(k-1)}.\end{align*}The series $\sum\limits_{k=1}^\infty\frac{1}{k(k-1)}$ converges, so it follows that $(f_n')_{n\in\mathbb{N}}$ converges uniformly on $[0,1]$. As $(f_n')_{n\in\mathbb{N}}$ converges uniformly we know that the sequence $(f_n)_{\in\mathbb{N}}$ must converge uniformly on $[0,1]$ as well (it is a theorem in our lecture).

This seems, if right at all, a bit of an overkill but I was not able to come up with a shorter or more elegant argument. Maybe someone knows a better way or can give me feedback if this correct.