Lower and upper bounds for $\tau(n)$

Question

How to prove the following statement:

If $n$ is the product of k powers of primes, i.e. $n=\prod\limits^{k}_{i=1}p_i^{\alpha_i}$ then $\omega (n) = k$ and $\Omega=\sum\limits_{i=1}^{k}\alpha_i$

$$ 2^{\omega (n)} \leq \tau(n) \leq 2^{\Omega(n)} $$

I'm stuck with this, not knowing where to begin.

Answer 1

The first two assertions,

• $\omega (n) = k$ and

• $\Omega(n) =\sum\limits_{i=1}^{k}\alpha_i$

are just the defintions.

Now recall that $\tau(n) = \prod_{i=1}^{k}(1+\alpha_i)$ and observe $$2 \le (1 + \alpha_i) \le 2^{\alpha_i}.$$