So I want to show that $$ \sum_f \lambda(f)t^{\deg f} = \prod_g \big(1 - \lambda(g)t^{\deg g}\big)^{-1} $$ where the sum is over monic polynomials and the product is over all monic irreducible polynomials in $F[x]$, where $F$ is a finite field.
Proof : So I know that this identity is proved by expanding each term $\big(1-\lambda(g)t^{\deg g }\big)^{-1}$ in a geometric series and using the fact that every monic polynomial can be written as a product of monic irreducible polynomials in a unique way. As you can see the details are left. Can you give me a complete proof?
So first expanding $\big(1-\lambda(g)t^{\deg g }\big)^{-1} $. We get $\displaystyle\sum_{r=0}^{\infty} \big(\lambda(g)t^{\deg g}\big)^r$. Now I have to show $$\displaystyle \sum_f \lambda(f)t^{\deg f} =\prod_g \left(\sum_{r=0}^{\infty} \big(\lambda(g)t^{\deg g}\big)^r\right)\,.$$ I only have to use the fact above. But how exactly? Attention definition of $\lambda$ : For a monic polynom $f(x) = x^n - c_1x^{n-1} + \cdots + (-1)^nc_n$ in $F[x]$ we define $\lambda(f) = \psi(c_1)\chi(c_n)$, where $\psi$ is a character with additive structure and $\chi$ is a multiplicative character. Moreover you should know that $\lambda$ is multiplicative. So $\lambda(fg) = \lambda(f) \lambda(g)$.