If $f(p^r) = c(c+1)^{r−1}$ for all prime powers $p^r$, then $F(n) = ∑_{d|n} f(d)$ is completely multiplicative. Let $c$ be any real number. Show that if $f$ is multiplicative such that
$f(1) = 1$ and $f(p^r) = c(c+1)^{r−1}$ for all prime powers $p^r$, then $F(n) = ∑_{d|n}
f(d)$ is completely multiplicative.
I know what it means to be a multiplicative function and to be completely multiplicative means it holds for all $n$ and not just when $gcd(n,m)=1$ I know how to prove that $F$ is multiplicative, but I'm not sure how to go about this or use the $f(p^r)$ identity. Any hints/solutions are greatly appreciated.
 A: Notice that, for prime $p$ and natural number $k$ $f(1)+f(p)+\dots+f(p^k)=1+c+\dots+c(c+1)^{k-1}=(c+1)^k$.
We have
\begin{align}
&F(d)=\sum_{h|d}f(h)=\sum_{h|p_1^{\alpha_1}p_2^{\alpha_2}\dots p_k^{\alpha_k}}f(h)=\sum_{h=p_1^{\beta_1}p_2^{\beta_2}\dots p_k^{\beta_k},~\beta_i\le\alpha_i}f(p_1^{\beta_1}p_2^{\beta_2}\dots p_k^{\beta_k})\\
=&\sum_{h=p_1^{\beta_1}p_2^{\beta_2}\dots p_k^{\beta_k},~\beta_i\le\alpha_i}\prod_{j=1}^kf(p_j^{\beta_j})=\sum_{\beta_1=0}^{\alpha_1}\sum_{\beta_2=1}^{\alpha_2}\dots\sum_{\beta_k=1}^{\alpha_k}\prod_{j=1}^kf(p_j^{\beta_j})\\
=&\sum_{\beta_1=0}^{\alpha_1}f(p_1^{\beta_1})\sum_{\beta_2=1}^{\alpha_2}f(p_2^{\beta_2})\dots\sum_{\beta_k=1}^{\alpha_k}f(p_k^{\beta_k})\\
=&\prod_{j=1}^k\sum_{\beta_j=0}^{\alpha_j}f(p_j^{\beta_j})=\prod_{j=1}^k(c+1)^{\alpha_j}=(c+1)^{\sum_{j=1}^k \alpha_j}
\end{align}
Which is $(c+1)$ to the number of prime divisors of $d$. So, for any $m$, $n$, $F(m)F(n)$ equals $c+1$ to the number of prime divisors of $m$ plus the number of prime divisors of $n$, where it is $c+1$ to the number of prime divisors of $mn$, which is $F(mn)$.
