Divisor Count Function from Prime Decomposition/Proof 2

Theorem
Let $n$ be an integer such that $n \ge 2$.

Let the prime decomposition of $n$ be:
 * $n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$

Let $\tau \left({n}\right)$ be the tau function of $n$.

Then:
 * $\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$

Proof
From Tau of Power of Prime we have:
 * $\forall j \in \left[{1 \,.\,.\, r}\right]: \tau \left({p_j^{k_j}}\right) = k_j + 1$

The result follows immediately Tau Function is Multiplicative‎.