For $R$-modules M, is $M\cong R^{\oplus n}\otimes_RM\cong M^{\oplus n}$? I wanted to explicitly give a bilinear map $R^{\oplus n}\times M\longrightarrow M^{\oplus n}$ when trying to prove that $R^{\oplus n}\otimes_RM\cong M^{\oplus n}$ and ended up with $(r_1,...,r_n,m)\mapsto ((\prod_{i=1}^nr_i)m,...,(\prod_{i=1}^nr_i)m)$ which seems to work in showing that every bilinear map to some $R$-module P : $R^{\oplus n}\times M\longrightarrow P$ factors through $M^{\oplus n}$. 
But I think the same works for $R^{\oplus n}\times M\longrightarrow M$, $(r_1,...,r_n,m)\mapsto (\prod_{i=1}^nr_i)m$. So is this true or did I do a mistake?
I'm in a commutative ring $R$ with identity. 
Edit: I probably should have made clear that I am trying to find a map that together with $M^{\oplus n}$ satisfies the universal property for the tensor product, and am not trying to give an isomorphism.
 A: The map you wrote is not an isomorphism. It is instead a map that arguably parametrizes (though not uniquely) all multiples of the diagonal map $M\rightarrow M^{\oplus n}$ (where the multiple, depending on an element $(r_1,\ldots,r_n)$ of $R^{\oplus n}$, is $\prod_ir_i$). This is why, as you observe, you could make a similar argument for $R^{\oplus n}\times M \rightarrow M$.
Instead, what you want is the map $(r_1,\ldots,r_n,m) \mapsto (r_1m,\ldots,r_nm)$. You can check that this induces the isomorphism $R^{\oplus n}\otimes M \rightarrow M^{\oplus n}$ that you want. If you know about the distributive property of the tensor product, you can also deduce this same result by noting the following:
$$
\begin{array}{rcl}
R^{\oplus n}\otimes M & = & (R \oplus \cdots \oplus R)\otimes M\\
& \cong & (R\otimes M)\oplus \cdots \oplus (R\otimes M)\\
& \cong & M\oplus \cdots \oplus M\\
& = & M^{\oplus n}
\end{array}
$$
A: It's much easier to show that if $A$, $B$ and $C$ are $R$-modules (commutative $R$, for simplicity), then
$$
(A\oplus B)\otimes_R C\cong (A\otimes_R C)\oplus (B\otimes_R C)
$$
The bilinear map $(A\oplus B)\times C$ is
$$
((a,b),c)\mapsto (a\otimes c,b\otimes c)
$$
and it's quite easy to show that this satisfies the universal property.
By induction, we get the isomorphism
$$
(A_1\oplus\dots\oplus A_n)\otimes_R C\cong
(A_1\otimes_R C)\oplus\dots\oplus(A_n\otimes_R C)
$$
In your case,
$$
R^{\oplus n}\otimes_R M\cong(R\otimes_R M)^{\oplus n}
$$
and you're done recalling that $R\otimes_R M\cong M$.
If you want the “explicit” bilinear map, it is indeed
$$
\varphi\colon((r_1,\dots,r_n),m)\mapsto (r_1m,\dots r_nm)
$$
Suppose you have a bilinear map $f\colon R^{\oplus n}\times M\to P$; define $f_i\colon M\to P$ by $f_i(m)=f(e_i,m)$, where $e_i$ is the usual basis element of $R^{\oplus n}$. Now define $g\colon M^{\oplus n}\to P$ by
$$
g(m_1,m_2,\dots,m_n)=f_1(m_1)+f_2(m_2)+\dots+f_n(m_n)
$$
Now check that $g\circ\varphi=f$.

Note that the map $R^{\oplus n}\times M\to M^{\oplus n}$ defined by
$$
((r_1,\dots,r_n),m)\mapsto 
\bigl(r_1r_2\dots r_n m,r_1r_2\dots r_n m,\dots,
r_1r_2\dots r_n m\bigr)
$$
is not bilinear (take $r_i=1+1$ and look).
The map $R^{\oplus n}\times M\to M$ defined by
$$
((r_1,\dots,r_n),m)\mapsto r_1r_2\dots r_nm
$$
is bilinear, but it fails to have the required universal property. Just think to the case $M=R$ and $n=2$: you would end up with $R^2\cong R$, which is not possible for commutative rings.
