Prove that $f(n) = 2^{\omega(n)}$ is multiplicative where $\omega(n)$ is the number of distinct primes 
Prove that $f(n) = 2^{\omega(n)}$ is multiplicative where $\omega(n)$ is the number of distinct primes.

My attempt:
Let $a = p_1p_2\cdots p_k$ and $b = q_1q_2\cdots q_t$
where $p_i$ and $q_j$ are prime factors, and $p_i \neq q_j$ for all $1 \leq i \leq k$ and $1 \leq j \leq t$.
We will show that $2^{\omega(ab)} = 2^{\omega(a)} \times 2^{\omega(b)}$
Indeed,
$\omega(a) = k$ and $\omega(b) = t$. Then $2^{\omega(a)} \times 2^{\omega(b)} = 2^{k + t}$
Where $2^{\omega(ab)} = 2^{k + t}$
$\therefore 2^{\omega(ab)} = 2^{\omega(a)} \times 2^{\omega(b)}$
Am I in the right track?   
Thanks,  

Note that if $f : \mathbb{N} \to \mathbb{C}$ is additive:
$$
\gcd(m,n) = 1 \quad \Longrightarrow \quad f(mn) = f(m) + f(n),
$$
then for any $t > 0$, $g(n) := t^{f(n)}$ is multiplicative:
$$
\gcd(m,n) = 1 \quad \Longrightarrow \quad g(mn) = g(m) g(n),
$$
since
$$
g(mn) = t^{f(mn)} = t^{f(m) + f(n)} = t^{f(m)}t^{f(n)} = g(m) g(n).
$$
Note also that this $g$ will never be the zero function.  Therefore, it suffices to show that $\omega(n)$ is additive (which has indeed already been shown).
