Here is an easier computation: $B_n$ admits a faithful representation into the automorphism group of a free group of rank $n$, $Aut(F_n)$—I learned this from a paper of Birman; the result might be older. Fix a free basis $x_1,\dotsc,x_n$. The image of the standard generator $\sigma_i$ is the automorphism sending $x_i$ to $x_ix_{i+1}x_i^{-1}$, sending $x_{i+1}$ to $x_i$, and fixing the other generators.

Conversely any automorphism permuting the conjugacy classes of the $x_i$ and fixing the word $x_1x_2\dotsb x_n$ is in the image of the representation.