Functional equation (N to N) Find all $f : \mathbb{N} \to \mathbb{N}$ which satisfy the equation:
$f(d_1)f(d_2)...f(d_n)=N$
Where $N$ is a natural number and $d_i, 1 \leq i \leq n$ are all of the divisors of $N$.

Define $f$ as follows: Let $$f(n) = \begin{cases} p & n=p^k \text{ for some prime } p \\ 1 & \text{otherwise (including $n = 1$)}\end{cases}$$.
Then, let $N = \displaystyle\prod_{i=1}^n p_i^{r_i}$. Then, note that while taking the product of $f$ of divisors of $N$, all terms have more than one prime will vanish. So, only left behind will be the following product:
$$
f(p_1) \ldots f(p_1^{r_1}) f(p_2) \ldots f(p_2^{r_2}) \ldots f(p_n) \ldots f(p_n^{r_n})
$$
But then, the value of each term is the prime inside, so we actually get this equal to $N$.
This also shows that the function is unique.
