$\Lambda_R [\alpha_1,...,\alpha_n]$ is the graded tensor product over $R$ of $\Lambda_R [\alpha_i]$ where the $\alpha_i$'s have odd dimension. In Example 3.16 of Hatcher's Algebraic Topology, it is written that the exterior algebra $\Lambda_R [\alpha_1,...,\alpha_n]$ is the graded tensor product over $R$ of the one-variable exterior algebras $\Lambda_R [\alpha_i]$ where the $\alpha_i$'s have odd dimension.
The definition of the exterior algebra is a given by:
The exterior algebra $\Lambda_R [\alpha_1,...,\alpha_n]$ over a commutative ring $R$ with identity is the free $R$-module with basis the finite products $\alpha_{i_1}...\alpha_{i_k}$ ($i_1<...<i_k)$, with associative, distributive multiplication defined by the rules $\alpha_i \alpha_j = - \alpha_j \alpha_i$ for $i \neq j$ and $\alpha_i^2=0$. The empty product of $\alpha_i$'s are allowed, and provides an identity element $1$ in $\Lambda_R[\alpha_1,...,\alpha_n]$. The exterior algebra becomes a (anti)commutative graded ring by specifying odd dimensions for the generators $\alpha_i$.
I understood the  definition of the exterior algebra, but I cannot understand the statement in Example 3.16. What is a "graded tensor product"?
Is it just meaning that $\Lambda_R[\alpha_1,...,\alpha_n]$ is isomorphic as $R$-algebras, to the tensor product $\Lambda_R[\alpha_1] \otimes _R .... \otimes _R \Lambda_R[\alpha_n]$?

Let $A=\bigoplus_{n=0}^\infty A_n$ and
$B=\bigoplus_{n=0}^\infty B_n$ be graded algebras over the ring $R$. We define
$C=A\otimes_R B$ as the vector space with grading
$$C_n=\bigoplus_{k=0}^n A_k\otimes_R B_{n-k}.$$
There are a couple of ways to make $C$ into a graded $R$ algebra. Here is the
less obvious way. Define, for $a\in A_k$, $b\in B_l$, $a'\in A_r$ and $b'\in B_s$,
$$(a\otimes b)(a'\otimes b')=(-1)^{lr}(aa')\otimes (bb')\in A_{k+r}\otimes_R
B_{l+s}\subseteq C_{k+l+r+s}.$$
If we define graded tensor product this way, then we can extend to products
of three or more graded algebras with no difficulty. If we consider 
$R[\alpha_i]$ where $\alpha_i^2=0$ and $\alpha_i$ is in dimension $1$,
then the sign rule enforces the relation $\alpha_i\alpha_j=-\alpha_j\alpha_i$
in the tensor product $R[\alpha_1]\otimes_R\cdots\otimes_RR[\alpha_n]$.
