Given $n + 1$ vector spaces $V_1, \ldots, V_{n + 1}$, a tensor product $\langle V_1 \otimes \cdots \otimes V_n, \otimes_n \rangle$ of $V_1, \ldots, V_n$ and a tensor product $\langle (V_1 \otimes \cdots \otimes V_n) \otimes V_{n + 1}, \otimes \rangle$ of $V_1 \otimes \cdots \otimes V_n, V_{n + 1}$, I want to prove that there exist a tensor product $\langle (V_1 \otimes \cdots \otimes V_n) \otimes V_{n + 1}, \otimes_{n + 1} \rangle$.
Here I use the following definition of a tensor product (see Finite Dimensional Multilinear Algebra from Marvin Marcus):
$\langle X_1 \otimes \cdots \otimes X_m, \otimes \rangle$ is a tensor product of $X_1, \ldots, X_m$ if:
$X_1 \otimes \cdots \otimes X_m$ is a vector space
$\otimes : \prod_{i \in \{ 1, \ldots, m \}} X_i \rightarrow X_1 \otimes \cdots \otimes X_m$ is a multilinear mapping
(universal property) If $\varphi : \prod_{i \in \{ 1, \ldots, m \}} X_i \rightarrow U$ is a multilinear mapping then there exists a unique linear mapping $h : X_1 \otimes \cdots \otimes X_m \rightarrow U$ such that $h \circ \otimes = \varphi$
$X_1 \otimes \cdots \otimes X_m = {span} \left( \otimes \left( \prod_{i \in \{ 1, \ldots, m \}} X_i \right) \right)$
Now for the proof:
$(V_1 \otimes \cdots \otimes V_n) \otimes V_{n + 1}$ is a vector space because $\langle (V_1 \otimes \cdots \otimes V_n) \otimes V_{n + 1}, \otimes \rangle$ is a tensor product.
Define $\otimes_{n + 1} : \prod_{i \in \{ 1, \ldots, n + 1 \}} V_i \rightarrow (V_1 \otimes \cdots \otimes V_n) \otimes V_{n + 1}$ by $\otimes_{n + 1} (v_1, \ldots, v_{n + 1}) = \otimes \left( \otimes_n (v_1, \ldots, v_n) {, v_{n + 1}} \right)$ then by the multilinearity of $\otimes$ and $\otimes_n$ it follows trivially that $\otimes_{n + 1}$ is multilinear.
(universal property) Here I'm stuck. How should I prove that, given a multilinear mapping $\varphi : \prod_{i \in \{ 1, \ldots, n + 1 \}} V_i \rightarrow U$, there exist a linear mapping $h : (V_1 \otimes \cdots \otimes V_n) \otimes V_{n + 1} \rightarrow U$ such that $h \circ \otimes_{n + 1} = \varphi$
Using the hints provided, I think that I have found the answer. Below is a sketch of the proof. I know that I can use the fact that $\langle V_1 \otimes \cdots \otimes V_n, \otimes_n \rangle$ is a tensor product and can do the following
Given a $v \in V_{n + 1}$ define $\varphi_v : \prod_{i \in \{ 1, \ldots, n \}} V_i \rightarrow U$ by $\varphi_v (v_1, \ldots, v_n) = \varphi (v_1, \ldots, v_n, v)$ which is clearly multilinear, hence there exist a linear map $k_v : (V_1 \otimes \cdots \otimes V_n) \rightarrow U$ such that $k_v \circ \otimes_n = \varphi_v$ because of the tensor product $\langle V_1 \otimes \cdots \otimes V_n, \otimes_n \rangle$.
We prove now that $k_v$ is linear in $v$. Let $u, v \in V_{n + 1}$, $\alpha \in F$ and $w \in \otimes_n \left( \prod_{i \in \{ 1, \ldots, n \}} V_i \right)$ then there exist a $(v_1, \ldots, v_n)$ such that $w = \otimes_n (v_1, \ldots, v_n)$. Further \begin{eqnarray*} k_{u + \alpha \cdot v} (w) & = & k_{u + \alpha \cdot v} (\otimes_n (v_1, \ldots, v_n))\\ & = & \varphi_{u + \alpha \cdot v} (v_1, \ldots, v_n)\\ & = & \varphi (v_1, \ldots, v_n, u + \alpha \cdot v)\\ & = & \varphi (v_1, \ldots, v_n, u) + \alpha \cdot \varphi (v_1, \ldots, v_n, v)\\ & = & \varphi_u (v_1, \ldots, v_n) + \alpha \cdot \varphi_v (v_1, \ldots, v_n)\\ & = & k_u (\otimes_n (v_1, \ldots, v_n)) + \alpha \cdot k_v (\otimes_n (v_1, \ldots, v_n))\\ & = & k_u (w) + \alpha \cdot k_v (w) \end{eqnarray*} proving that $k_{u + \alpha \cdot v} = k_u + \alpha \cdot k_v$.
Define $\psi : (V_1 \otimes_n \cdots \otimes_n V_n) \cdot V_{n + 1} \rightarrow U \text{ by } \psi (u, v) = k_v (u)$ then we have $\psi (x + \alpha \cdot y, v) = k_v (x + \alpha \cdot y) = k_v (x) + \alpha \cdot k_v (y) = \psi (x) + \alpha \cdot \psi (y)$ and $\psi (v, x + \alpha \cdot y) = k_{x + \alpha \cdot y} (v) = k_x (v) + \alpha \cdot k_v (v) = \psi (v, x) + \alpha \cdot \psi (v, y)$
From the above it follows that \ $\psi$ is multilinear. As $\langle (V_1 \otimes_n \cdots \otimes_n V_n) \otimes V_{n + 1}, \otimes \rangle$ is a tensor space there exist a linear mapping $h : (V_1 \otimes \cdots \otimes V_n) \otimes V_{n + 1} \rightarrow U$ such that \ $h \circ \otimes = \psi$. Let $(v_1, \ldots, v_{n + 1}) \in \prod_{i \in \{ 1, \ldots, n + 1 \}} V_i$ then we have \begin{eqnarray*} (h \circ \otimes_{n + 1}) (v_1, \ldots, v_{n + 1}) & = & h (\otimes_{n + 1} (v_1, \ldots, v_{n + 1}))\\ & = & h (\otimes (\otimes_n (v_1, \ldots, v_n), v_{n + 1}))\\ & = & \psi (\otimes_n (v_1, \ldots, v_n), v_{n + 1})\\ & = & k_{v_{n + 1}} (\otimes_n (v_1, \ldots, v_n))\\ & = & \varphi_{v_{n + 1}} (v_1, \ldots, v_n)\\ & = & \varphi (v_1, \ldots, v_n, v_{n + 1}) \end{eqnarray*} which proves that \ $h \circ \otimes_{n + 1} = \varphi$.
It is a little bit more complex as in this text because we have to work with spans but I can solve this myself.
Thanks for the help.