For a commutative ring $R$ and an $R$-module $M$, define the exterior algebra $\bigwedge M$ as the quotient algebra $T(M)/J$ with $J$ being the two-sided ideal of $T(M)$ generated by the elements of the form $m\otimes m, m \in M$. Set $\bigwedge^n M = (\bigotimes^n M + J)/J$.
Let $\pi_I\colon T(M)\to T(M)/J$ be the canonical projection and $l_n\colon \bigotimes^n M \to \bigoplus_{n \geq 0} \bigotimes^n M = T(M)$ be the canonical injection. When $\mathrm{im}(\pi_I\circ l_n) = \bigwedge^n M$, so we may consider a canonical $R$-linear map $p_n\colon \bigotimes^n M\to \bigwedge^n M$ given by $(m_1,...,m_n) \mapsto m_1\wedge ... \wedge m_n$.
I want to prove that the kernel of this map is generated by $m_1\otimes ... \otimes m \otimes m \otimes ... \otimes m_n$ as a su bmodule of $\otimes^n M$. One direction is trivial, but I'm not sure how to prove that $\mathrm{ker}(p_n)$ is a subset of this submodule. Let $\sum_{i = 1}^k m_{i,1}\otimes ... \otimes m_{i,n}$ be a typical element of $\bigotimes^n M$. If it belongs to $\ker(p_n)$, then $\sum_{i = 1}^k m_{i,1}\otimes ... \otimes m_{i,n} \in J$. But it doesn't follow that $m_{i,1}\otimes ... \otimes m_{i,n} \in J$ for $i = 1,...,k$.