I am looking to verify my answer to the question $$F(n)=\sum_{d|n}{\mu(d)\sigma(d)}=(-1)^{\omega(n)}\prod_{j=1}^{\omega(n)}{p_j}$$ Where $\mu$ is the Möbius function, $\sigma$ is the sum of divisors of a number $n$, and $\omega$ is the number of distinct prime factors of $n$.
Let $n=\prod_{j=1}^kp_j^{\alpha_j}$. Now, since the function $\mu(n)\sigma{(n)}$ is multiplicative, so is $F(n)$ and thus I can look at $F(p_i^{\alpha_i})$ Now the divisors of $p_i^{\alpha_i}$ are $1, p_i, p_i^2, ..., p_i^{\alpha_i}$, so $\mu(p_i^{\alpha_i})=0$ if $\alpha_i>1$. Therefore, for each prime power factor of $n$, we get that $$F(p_i^{\alpha_i})=\mu(1)\sigma(1)+\mu(p_i)\sigma(p_i)$$ $$=1\cdot 1 + (-1)\cdot (p_i+1)=-p_i$$ Since there are $\omega(n)$ primes, the formula above follows.
Is this correct? My gut says yes...
Your proof is correct.
If there is anything to remark, it is that your proof only covers the case where $n$ has at least one prime divisor, that is, $n>1$. This means one should verify the case $n=1$ separately. When $n=1$ the product is empty and hence (by convention) equal to $1$, which is the same as the LHS.