Find the strictly increasing functions $f: \mathbb{N}^{*} \rightarrow \mathbb{N}^{*}$ such that $f(2)=2$ and if $n, m \geq 1$,

$$
f(n m)=f(n) f(m)
$$

(MEMO 2014)
We immediately have $f(1)=1$. Let $f(3)=a$, and we aim to show that $a=3$. $f(6)=2a$, so $a+2 \leq f(5) \leq 2a-1$ by strict monotonicity, and on one hand $f(10) \leq 4a-2$, hence $f(9) \leq 4a-3$ and $f(18) \leq 8a-6$. On the other hand, $f(15) \geq a(a+2)$.

Thus:

$$
a(a+2) \geq f(15) \geq f(18)-3 \geq 8a-9
$$

or equivalently $(a-3)^2 \leq 0$, leading to the result.

Let's note the following two facts: if $f(n)=n$ and $f(n-1)=n-1$, then $f(n(n-1))=n(n-1)$ (since $\gcd(n, n-1)=1$), and if for some $k \geq 1, f(k)=k$, then for all $j \geq k, f(j)=j$ (by strict monotonicity). These allow us to easily prove by induction on $n$ that $f(n)=n$ for all $n \in \mathbb{N}^{*}$, with $f(3)$ being necessary for initialization. Thus, the identity is the only solution to the problem.