There are $n$ numbers $a_1, a_2, ..., a_n$. Two players take turns. In each move, player chooses some number $a_i$ and writes it as a product of two numbers in any way he wants (so for exapmle $1470 = 98 * 15$ or $1470 = 14 * 105$). Let's say he chose $1470 = 98 * 15$, then he replaces $1470$ with $15$'s writen $98$ times and the game continues. Player loses if he can't do any move (all numbers are equal to $1$).
Example game: $(2, 9) - (2, 3, 3, 3) - (2, 3, 1, 1, 1, 3) - (1, 1, 3, 1 1 1, 3) -$ $(1, 1, 1, 1, 1, 1, 1, 1, 3) - (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)$ and the second player loses.
How to tell which player will win if both play optimally? I think this game can be decomposed into $n$ smaller games on $a_1$, $a_2$, ..., $a_n$ using Sprague-Grundy theorem. I've tried to reduce smaller games to NIM, but with no effect.
Claim: each number is a nim-heap. The number stones in the heap is the sum of the exponents of odd prime factors plus $1$ if the number is even. Your example of $1470=2\cdot 3 \cdot 5 \cdot 7^2$ corresponds to a heap of $5$. $2940=2^2\cdot 3 \cdot 5 \cdot 7^2$ would also be a heap of $5$.
To prove the claim, first note that any prime including $2$ is a nim-heap of $1$ because you can only move to an empty heap. The product of two odd primes then is a nim-heap of $2$ because you can move to $0$ by splitting into $1$s or into $1$ by splitting into one of the primes. The product of three odd primes is a nim-heap of $3$ because you can split into $0,1,2$ and so on. Powers of $2$ do not increase the heap because if you split one of them you get an even number of matching heaps which is equivalent to a heap of $0$. We just do induction on the number of odd factors to prove the claim.