Construction of tensor algebra: extending the multiplication from being defined on pure tensors T.S.Blyth in his book Module Theory: An Approach to Linear Algebra defined multiplication on the tensor algebra $\bigotimes M = \bigoplus_{n \in \mathbb{N}}\bigotimes^n M$ by first defining it for pure tensors:
$$1_R(x_1\otimes ... \otimes x_n) = x_1\otimes ... \otimes x_n = (x_1\otimes ... \otimes x_n)1_R$$
 and 
$$(x_1\otimes ... \otimes x_n)(y_1\otimes ... \otimes y_m) = x_1\otimes ... \otimes x_n \otimes y_1 \otimes ... \otimes y_m ~ ,$$
where $$x_1\otimes ... \otimes x_n \in \bigotimes^n M ~~ \text{and} ~~ y_1\otimes ... \otimes y_m \in \bigotimes^m M~.$$ He then says that it is possible to extend this definition to multiply arbitrary elements $\bigotimes^n M$ and $\bigotimes^m M$ since every elements of $\bigotimes^n M$ is a linear combination of pure tensors and the same is true for elements of $\bigotimes^m M$. This, is $x = \sum_i r_ix_i$ and $y = \sum_i s_iy_i$ where $x_i \in \bigotimes^n M$ and $y_i \in \bigotimes^m M$ are pure tensors, that $xy = \sum_{i,j} (r_is_j)(x_iy_j)$ is well-defined, that is, it doesn't depend on the choice of linear combinations of $x$ and $y$. However, the latter is not clear to me.
How to prove that is $$\sum_i r_ix_i = \sum_i r'_ix'_i ~~ \text{and} ~~ \sum_i s_iy_i = \sum_i s'_iy'_i, $$ 
then 
$$\sum_{i,j} (r_is_j)(x_iy_j) = \sum_{i,j} (r'_is'_j)(x'_iy'_j)$$ 
where $x_i,x'_i \in \bigotimes^n M$ and $y_i,y'_i \in \bigotimes^m M$ are pure tensors?
 A: The multiplication map is just a collection of minor reinterpretations of the tensor maps that are already attached to the tensor products. The tensor map
$$\bigotimes^n M × \bigotimes^m M → \bigotimes^n M \otimes \bigotimes^m M, (x,y) ↦ x \otimes y$$
is well-defined $A$-bilinear, thus by associativity of the tensor product, it defines a well-defined map
$$\bigotimes^n M × \bigotimes^m M → \bigotimes^{n+m} M,$$
which exactly is the multiplication of tensors of degree $n$ and $m$.
Using the universal property of direct summands and inclusions $\bigotimes^{n+m} M → \bigotimes M$, we can put all these maps together to give the multiplication
$$\bigotimes M × \bigotimes M → \bigotimes M.$$

For any three $A$-modules $M_1, M_2, M_3$, there is an associator map
$$(M_1 \otimes M_2) \otimes M_3 → M_1 \otimes (M_2 \otimes M_3),~(x\otimes y) \otimes z ↦ x \otimes (y \otimes z)$$
which is an isomorphism. Whether you define $\bigotimes^n M$


*

*directly, or

*inductively as $M \otimes \bigotimes M^{n-1}$, or

*inductively as $\bigotimes M^{n-1} \otimes M$,


these associator maps are the building blocks to define natural isomorphisms for all versions of $l$-ary tensor products. In particular $\bigotimes^n M \otimes \bigotimes^m M = \bigotimes^{n+m} M$, canonically.
To actually formally prove this, we would either directly dig through inductions and calculations about very formal statements or device some simple formal machinery to manage these inductions and calculations more easily.
But nobody does that. Usually, one just sees that associativity as we know it holds and carries on.
A: To the answer of k.stm I will add excellent notes of Daniel Murfet which adress the matter rigorously: 
Tensor, Symmetric and Exterior Algebras
They include a detailed construction of a canonical isomorphism $\bigotimes^n M\otimes\bigotimes^k M\to \bigotimes^{n + k} M$ which k.stm referred to. Associativity of the multiplication in $\bigotimes M$ is also treated.
