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,  

Hint: Notice $$2^{\omega(a)}\times 2^{\omega(b)}=2^{\omega(a)+\omega(b)}$$ so try to relate $\omega(a)+\omega(b)$ and $\omega(ab)$.
To simplify things, you need only show it holds for $a=p^r$, $b=q^t$ where $q,p$ are prime numbers. 
