Let $V$ be a $k$-vector space and $V^{\otimes n}$ the $n$-fold tensor power of $V$ and let $\mathbb{S}_n$ be the symmetric group of an n-element set, with its signum representation denoted by $(-1)^\sigma$ for $\sigma\in \mathbb{S}_n$.
Now on one side we have the "usual" symmetric tensor power $V^{\odot n}$ of $V$ defined by quotienting the ordinary tensor power by the maximal ideal generated by "sums over antisymmetriezed permutations", i.e.:
$V^{\odot n}:= V^{\otimes n} / \langle\{\sum_{\sigma\in \mathbb{S}_n} (-1)^\sigma v_{\sigma(1)}\otimes...\otimes v_{\sigma(n)}\;|\; v_1\otimes...\otimes v_n \in V^{\otimes n}\}\rangle$
On the other side it is said, that the same symmetric tensor power can be obtained as the following tensor product:
$V^{\otimes n} \otimes_{k[\mathbb{S}_n]} k$
where $k[\mathbb{S}_n]$ is the group algebra over $\mathbb{S}_n$, with $k$ a left $\mathbb{S}_n$-module, induced from the trivial representation and with $V^{\otimes n}$ a right $\mathbb{S}_n$ module induced from the right representation of $\mathbb{S}_n$ which permutes the indices.
Can someone explain, how $V^{\otimes n} \otimes_{k[\mathbb{S}_n]} k$ is equivalent to the symmetric tensor product $V^{\odot n}$? I'm not sure, how we should think about $V^{\otimes n} \otimes_{k[\mathbb{S}_n]} k$ or tensor products $\bullet \otimes_{k[\mathbb{S}_n]} \bullet$ in general.
For any $k[\mathbb{S}_n]$-module $M$, $M\otimes_{k[\mathbb{S}_n]}k$ is the largest quotient of $M$ on which $\mathbb{S_n}$ acts trivially.
Factoring $V^{\otimes n}$ by the ideal generated by differences of elements that differ just by swapping two tensor factors gives the largest quotient on which transpositions act trivially.
So it comes down to the fact that $\mathbb{S}_n$ is generated by transpositions.