Showing that multilinear maps factor through tensor products Let $R$ be a commutative ring and let $M_{1},...M_{n},L$ be $R$-modules.  Let $\iota:M_{1},...M_{n}\rightarrow M_{1}\bigotimes...\bigotimes M_{n}$ be defined by $\iota(m_{1},...,m_{n})\mapsto m_{1}\otimes...\otimes m_{n}$ and let $\phi:M_{1}\times...\times M_{n}\rightarrow L$ be an $n$-multilinear map.  Claim: there is a unique $R$-module homomorphism $\Phi:M_{1}\bigotimes...\bigotimes M_{n}\mapsto L$  such that $\phi =\Phi \circ \iota$.  

Assume this holds for $n=2$ (in other words, assume the universal property for the tensor product of pairs of modules).  I want to show that this case implies the remaining cases, for example, $n=3$.  Apparently this is immediate, but I have absolutely no idea why it is true.  I have tried every way of pairing the modules but nothing has worked.  

Edit: I am using the construction from free groups, etc. definition of tensor products.
Existence

$\Phi: m_1 \otimes \cdots \otimes m_n \mapsto \phi(m_1, \cdots, m_n)$ satisfies the condition.

Uniqueness

Let $\Phi$ and $\Phi'$ be maps satisfying the condition.

Then, I will prove that $\Phi = \Phi'$.

Let $m_1 \otimes \cdots \otimes m_n \in M_1 \otimes \cdots \otimes M_n$. Then:

$$\begin{array}{rcl}
\Phi(m_1 \otimes \cdots \otimes m_n)
&amp;=&amp; \Phi(\iota(m_1,\cdots,m_n)) \\
&amp;=&amp; \phi(m_1,\cdots,m_n) \\
&amp;=&amp; \Phi'(\iota(m_1,\cdots,m_n)) \\
&amp;=&amp; \Phi'(m_1 \otimes \cdots \otimes m_n)
\end{array}$$
Existence

$\Phi: m_1 \otimes \cdots \otimes m_n \mapsto \phi(m_1, \cdots, m_n)$ satisfies the condition.

Uniqueness

Let $\Phi$ and $\Phi'$ be maps satisfying the condition.

Then, I will prove that $\Phi = \Phi'$.

Let $m_1 \otimes \cdots \otimes m_n \in M_1 \otimes \cdots \otimes M_n$. Then:

$$\begin{array}{rcl}
\Phi(m_1 \otimes \cdots \otimes m_n)
&amp;=&amp; \Phi(\iota(m_1,\cdots,m_n)) \\
&amp;=&amp; \phi(m_1,\cdots,m_n) \\
&amp;=&amp; \Phi'(\iota(m_1,\cdots,m_n)) \\
&amp;=&amp; \Phi'(m_1 \otimes \cdots \otimes m_n)
\end{array}$$