So by Fundamental Arithmetic Theorem, any integer has a unique prime factorization into primes, written as: $$n=p_1^{k_1}p_2^{k_2}p_3^{k_3}...p_r^{k_r}$$ From exponents $k_1,...k_r$ it is possible to find the number of divisors of $n$, called divisor function written as $d(n)$. Find $d(100)$ and $d(320)$. Then, find a sgeneral formula for $d(n)$ in terms of $k_1,...,k_r.$
Having trouble figuring out where to start here. Any guidance would be really great! Thank you.
Hint
A divisor of $n$ can be written (uniquely) in the form $$ n=p_{1}^{\alpha_1}p_{2}^{\alpha_2}\dotsb p_{r}^{\alpha_r} $$ where $0\leq \alpha_i\leq k_{i}$. How many choices are there for $\alpha_1$, for $\alpha_2$, and so on? Then use the multiplication principle to get the result.