A free direct sum of a projective module 
I want to prove that a left module $_RP$ is a projective generator if and only if a direct sum of (copies of) $P$ is free. 

My try is, first, to observe that $P$ is a generator if and only if for some positive integer $n$ and some left $R$-module $X$ we have $P^{(n)}$ is isomorphic to $R⊕X$. Also, $P$ is a direct summand of a free $R$-module. Henceforth I am stuck. Any help would be appreciated.
 A: 
Claim: If $P,Q$ are two projective generators of $R\text{-Mod}$, then there exist sets $I,J$ such that $P^{(I)}\cong Q^{(J)}$

Proof: You can use the Eilenberg-Mazur Swindle here: Suppose $P,Q$ are projective generators. Then, to begin, using the generator property of $P$, there exists a surjection $P^{(I_0)}\twoheadrightarrow X_0:=Q$. By projectivity of $Q$, this surjection splits, i.e. $P^{(I_0)}\cong X_0\oplus Y_0$ for some $Y_0$. As $P$ is projective, so is $Y_0$. Next - this time using the generator property of $Q$ - we find a surjection $Q^{(J_0)}\twoheadrightarrow Y_0$, which, by projectivity of $Y_0$, splits again, i.e. $Q^{(J_0)}\cong Y_0\oplus X_1$, with $X_1$ projective again since $Q$ is projective. Now continue this procedure, producing sets $I_n$ and $J_n$ and isomorphisms $P^{(I_n)}\cong X_n\oplus Y_n$ and $Q^{(J_n)}\cong Y_n\oplus X_{n+1}$. Then $$\bigoplus_n P^{(I_n)}\cong X_0\oplus\bigoplus_n Q^{(J_n)} = Q\oplus\bigoplus_n Q^{(J_n)}$$
showing that suitable direct-sum-powers of $P$ and $Q$ are isomorphic. $\quad\square$
Your statement follows by taking $P$ some projective generator and $Q := R$.
