I searched and wasn't able to find a question similar enough to mine. Here's the problem:
Show that $\sigma=c_1*c_1$, where $c_1$ is the constant function $1$.
Here is my attempt. My argument makes sense to me, but it seems kind of short. I'm wondering if there is anything I need to add to my argument or if I'm sort of using circular reasoning.
The operation $*$ is the convolution product defined as: \begin{equation} \begin{split} (c_1 * c_1)(n) &= \sum_{d \mid n}c_1(d)c_1\left(\frac nd \right) \\ \end{split} \end{equation}
The formula above is clearly true if $n=1$. Assume that $n > 1$ and write $n=p_1^{e_1} \dots p_s^{e_s}$. In this sum, all terms, $c_1(d)c_1(\frac nd)$, are equal to $1$ for all $d \mid n$. So we will end up multiplying $1$ by the number of divisors, which is exactly $\sigma(n)$.
As always, thank you all for your help.