I denote by $m_n(q)$ the number of irreducible monic polynomials with degree $n$ over the finite field of order $q$. So the number of monic polynomials with degree is just $q^n$.
From this, how does it follow that $$ \prod_{n\geq 1}\frac{1}{(1-x^n)^{m_n(q)}}=\frac{1}{1-qx}. $$ I know $$ \frac{1}{1-qx}=1+qx+q^2x^2+\cdots $$ but I don't see an obvious relation between the coefficients of one to the other.
Thanks,
2nd try: Based on the help, I try to write this out in form $$ (1+x+x^2+\cdots)^{m_1(q)}(1+x^2+x^4+c\dots)^{m_2(q)}\cdots=\sum_{f\text{ monic}}x^{\deg f}. $$
Due to unique factorization, distinct products of monic irreducibles of determine unique monic polynomials. So the coefficient of $x^{\deg f}$ is the number of monic polynomials of degree $\deg f$, But I'm having a hard time picturing how the expanded left hand side would count them for the identity to hold. Would someone be willing to explain this subtlety?
$q^n$ is the number of monic polynomials of degree $n$ over $\mathbb{F}_q$, so we can write $\frac{1}{1 - qx}$ as $$\sum_{f \text{ a monic polynomial}} x^{\deg f}.$$
Now, monic polynomials can be uniquely factored into a product of irreducible polynomials, and $\deg$ satisfies $\deg fg = \deg f + \deg g$. What does that tell you about how the above generating function can be factored? (This is closely analogous to how we derive the Euler product for the Riemann zeta function.)