I need to show that the abelianization of braid group $B_n$, for $n\geq2$ is isomorphic to $\mathbb{Z}$, and that the commutator subgroup $[B_n,B_n]$ is exactly the set of braids represented by words with total exponent sum zero in the generators $\sigma_{i}$. I was able to show all generators $\sigma_i$ are conjugate to each other.
Here is the question:
The fact that all generators are conjugate (which you have shown) implies that all generators have the same image in $G/[G,G]$ (where $G=B_n$), so $G/[G,G]$ is cyclic and generated by any one of the generators.
On the other hand, the map $\phi$ sending every generator $\sigma_i$ to $1 \in {\mathbb Z}$ induces a surjective homomorphism, so $G/[G,G]$ is infinite, and hence infinite cyclic.
Note that $\ker \phi = [G,G]$ is the set of elements with total exponent sum $0$ in the generators.