Cohomology groups of the bar construction of free graded algebras Let $A= \bigwedge \langle x_n \rangle $ be the free  graded commutative algebra with differential zero ($x_n$'s are elements with positive grading )over a field of characteristic zero. I want a reference for computation of cohomology ring of $\bar{\mathrm{B}}^\bullet(A)$, reduced bar complex of $A$. It seems it's equal to $\bigwedge \langle x_n[1] \rangle $ but I can't do the computations.
Thanks!
 A: It is done in the Cartan Seminar:

Séminaire Henri Cartan de l'École Normale Supérieure, 1954/1955. Algèbres d'Eilenberg–MacLane et homotopie. 2ème éd. Secrétariat mathématique, 11 rue Pierre Curie, Paris, 1956. iii+234 pp. MR0087935

You can find the first part and the second part on Numdam. I think the precise reference is either the 3rd or 4th talk. If you do not read French or don't want to translate into modern language (they use the terminology of "multiplicative construction" of which the reduced bar complex is a special case), it follows from the Hochschild–Kostant–Rosenberg theorem, see e.g. Theorem 9.4.7 of

Charles A. Weibel. An introduction to homological algebra. Cambridge Studies in Advanced Mathematics 38. Cambridge: Cambridge University Press, 1994, pp. xiv+450. ISBN: 0-521-43500-5; 0-521-55987-1. MR1269324.

(because the homology of the reduced bar complex is the Hochschild homology with trivial coefficients $H_*(\bar{B}A) \cong HH_*(A; \Bbbk)$). The precise result is indeed that if $A$ is a symmetric graded algebra on generators $x_i$, then $HH_*(A;\Bbbk)$ is isomorphic as a graded vector space to the symmetric graded algebra on the generators $x_i[1]$ with degree shifted by $1$. Another possible reference is Proposition 6.2 of

Benoit Fresse. “Iterated bar complexes of E -infinity algebras and homology theories”. In: Algebr. Geom. Topol. 11.2 (2011), pp. 747–838. ISSN: 1472-2747. DOI: 10.2140/agt.2011.11.747. MR2782544.

though it is written in the language of operads; but I don't think there are major difficulties involved in recovering the original arguments for a plain algebra (or using $\mathsf{C}(\Sigma I) \circ V = S(\Sigma V) \xrightarrow{\sim} \operatorname{B}(\mathsf{C}(I)) \circ V \cong \operatorname{B}(S(V))$ because all the assumptions are satisfied).
