In a paper I found the following result: $$\mathrm{GL}_n(\mathbb{Z}_2)=\mathrm{Out}(F_n)/\langle\langle \epsilon_1,\dots,\epsilon_n\rangle\rangle$$ However, they got the result as a corollary of a pretty strong result. I was wondering if there is a direct proof for this fact. Here $\epsilon_i$ is the automorphism of $F_n$ that inverts the $i$-th generator and leaves the rest fixed.

Thanks for your help.

Edit: The natural approach I was thinking is to consider the map given by the abelianization and compose it with the reduction mod $2$: $$\varphi:\mathrm{Out}(F_n)\to\mathrm{GL}_n(\mathbb{Z})\to\mathrm{GL}_n(\mathbb{Z}_2)$$ This is trivially an epimorphism and I proving that $\ker(\varphi)=\langle\langle \epsilon_1,\dots,\epsilon_n\rangle\rangle$ will solve the problem. $\mathrm{Out}(F_n)$ is generated by $\epsilon_1,\dots,\epsilon_n,\lambda_{i,j}$ with $i,j=1,\dots,n$ and $i\neq j$. ($\lambda_{i,j}(v_i)=v_iv_j$ and $\lambda_{i,j}(v_k)=v_k$ if $k\neq i$, where $v_1,\cdots,v_n$ are the generators of $F_n$). But there are elements such as $\lambda_{i,j}^2$ that lie in $\ker(\varphi)$ and I cannot prove that they lie in $\langle\langle \epsilon_1,\dots,\epsilon_n\rangle\rangle$.