Let $V$ be a (co)chain complex, and let $Sym(V)$ be the free differential graded-commutative algebra generated by $V$. Definition and examples below in case you don't know what I mean.
Question: What's the cleanest proof that $H^*(Sym(V)) \cong Sym(H^*(V))$?
All my proofs are a little messy, using the gradings on Sym(V) coming from cohomological grading and degree by the number of products. Though I expect all proofs to use this bigrading, a clean, succinct proof would be very appreciated!
Definition: $Sym(V)$ is the free dg tensor algebra on $V$, modded out by the relation $v_1 v_2 = (-1)^{|v_1| |v_2|} v_2 v_1$. The differential is given by the Leibniz rule: $d(uv) = (du)v + (-1)^{|u|}u dv$.
Example: If $V$ is concentrated in degree 0, $Sym(V)$ is the usual free commutative algebra generated by $V$. If $V$ is concentrated in even degrees, the underlying vector space of $Sym(V)$ is the usual free commutative algebra, but $Sym(V)$ has a grading where $|v_1 \ldots v_n| = |v_1| + \ldots + |v_n|$. The differential is zero by the Leibniz rule.
Example: If $V$ is concentrated in odd degrees, $Sym(V)$ has underlying vector space given by the exterior algebra generated by $V$. It has no differentials by the Leibniz rule, but has grading $|v_1 \ldots v_n| = |v_1| + \ldots + |v_n|$.
This requires some sort of assumption on the ground ring, or on $V$ or both. Consider, for example, the ground ring $\mathbb Z/2$ and the complex $V = \{a\to da\}$ where $a$ has degree $n>1$. Then, clearly $Sym(H^*(V))$ is $\mathbb Z/2$ in degree zero. But, $a^2$ represents a non-zero class in $H^{2n}(Sym(V))$. You should assume that the ground field has characteristic zero.
Edit: By induction you can prove the result by considering the filtration $Sym^n(V) = \bigoplus_{i=0}^n V^{\otimes i}/\Sigma_i$. Then, the quotient $Sym^{n+1}(V)/Sym^n(V) = V^{\otimes (n+1)}/\Sigma_{n+1}$, and $\mathbb k$ is a projective $\mathbb k [\Sigma_{n+1}]$ module, thus $H^*(V^{\otimes (n+1)}/\Sigma_{n+1}) \cong H^*(V)^{\otimes (n+1)}/\Sigma_{n+1}$. Now induction tells you that $H^*(Sym^n(V)) \cong Sym^n(H^*(V))$, and thus the result.