Natural Isomorphism of $S(V[1])$ and $(\bigwedge V)[n]$ Let $V:=\oplus_{j\in\mathbb{Z}}V_j$ be a graded $\mathbb{F}$-vector space over
the field $\mathbb{F}$. The graded tensor product of graded vector spaces is given
by 
$V \otimes W:= \oplus_{j\in \mathbb{Z}}\oplus_{p+q=j}V_p\otimes V_q$
and for the graded vector space $\mathbb{F}[j]$, which is $\mathbb{F}$ in degree
$j$ and te zero vector space $\{0\}$ otherwise, the shift $V[j]$ is given by
$V[j]:=\mathbb{F}[j]\otimes V$
We then define a monoidal structure on the category of graded vector space,
(more or less) given by the rule on homogeneous elements 
$v\otimes w= (-1)^{deg(v)deg(w)}w\otimes v$
Then there is the decalage isomorphism
$
dec: V_1[1]\otimes \cdots \otimes V_n[1] \to (V_1 \otimes \cdots \otimes V_n)[n]
$
given by $dec(v_1[1]\otimes \cdots \otimes v_n[1])= 
(-1)^{\sum_{j=1}^n(n-j)deg(v_j)}(v_1\otimes \cdots \otimes v_n)[n]$.
Now in work on graded (stuff), it is frequently said, that this isomorphism
defines a natural isomorphism of the symmetric graded tensor-algebra of $V[1]$
and the antisymmetric graded tensor algebra, that is
$S(V[1])\simeq (\bigwedge V)[n]$
*The question is: How does the decalage induces such an algebra isomorphism? Or
What is the natural isomorphism? *
If $dec$ itself would be the isomorphism, then
$dec(v[1] \vee w[1])= (dec(v_1)\wedge dec(w))[2]$ should hold, but this isn't true in general.
 A: You have a detailed proof (in a much more general context but easy to read) in Proposition I.4.3.2.1 of Illusie, Complexe Cotangent et Déformations I, Springer LNM 239. 
A: Let $V$ be a $\mathbb Z$ graded vector space over a field $\mathbb k$ of characteristic $0$ (for simplicity). The suspension $V[1]$ of $V$ is the graded vector space
$V[1]:= V\otimes_{\mathbb k} \mathbb k[1],$ with $\mathbb k[1]$ concentrated in degree $-1$, and $\mathbb k_{-1}=\mathbb k$. With $s$ we denote the "suspension" morphism $s:V\rightarrow V[1]$, $x\mapsto sx:=x$ of degree $-1$; in other words, $|sx|=|x|-1$, where $|\cdot|$ denotes the degree of any homogeneous element in $V$ (or any other graded vector space). In general, we write $s^n: V\rightarrow V[n]$, for any integer $n$. We introduce the graded symmetric resp. antisymmetric algebras over $V$:
$$S(V ) = T (V )/ \langle x \otimes y − (−1)^{|x||y|}
y \otimes x \rangle,~~ \text{resp.}~~\Lambda (V ) = T (V )/ \langle x \otimes y + (−1)^{|x||y|} y \otimes x \rangle, $$
For any $n\geq 0$ the decalage is a canonical isomorphism of graded vector spaces (not of algebras)
$$\Phi_n: S_n(V[1] )\rightarrow \Lambda(V)[n], $$
where $$\Phi_n(sx_1,\cdots, sx_n):= (-1)^{\sum_{i=1}^n(n-i)|sx_i|}s^{n}(x_1\wedge\cdots\wedge x_n). $$
Why that sign? The sign follows from the Koszul rule, once one removes the suspensions $s$ from the string $sx_1,\cdots, sx_n$ one by one applying $n$-times the desuspension morphism $s^{-1}$.
