Product of ideals for Nakayama's Lemma The result to be proved is the following:
Let $R$ be a local Noetherian ring. Then the minimum number of generators of the unique maximal ideal $P$ equals the dimension of $P/P^2$ as a vector space over $R/P$.
Proof:
$P$ is generated by $x_1, ... , x_n$ $\iff$ $P/P^2$ is generated by $\overline{x_1}, ... , \overline{x_n}$, where a bar indicates the image of an element in the quotient $P/P^2$.
($\implies$) Clear.
($\impliedby$) Suppose $\overline{x_1}, ... , \overline{x_n}$ generate $P/P^2$. Consider $I=(x_1, ... , x_n) \leq P$ . Then $I + P^2 =P$. 
$\mathbf{Hence}$, $\mathbf{P(P/I)=P/I}$.
Then Nakayama implies $P/I=0$ and we're done. 
My question: I am completely stuck on the part in bold. I will appreciate explicit/trivial answers most. Thank you!
 A: Let $M$ be a (left) module and $L$ a submodule of $M$. Let $I$ be a (left) ideal of $R$ (no assumption on the ring). Then $IM$ is a submodule of $M$ and
$$
I(M/L)=(IM+L)/L
$$
Proof. The submodule $I(M/L)$ of $M/L$ is generated by the elements of the form $r(x+L)$ for $r\in I$ and $x\in M$. Such an element is in $(IM+L)/L$. Conversely, an element of $(IM+L)/L$ is of the form
$$
\sum_{k=1}^n ((r_kx_k+l_k)+L)=\sum_{k=1}^n (r_kx_k+L)\in I(M/L).\tag*{QED}
$$
Thus, if you have proved that $P^2+I=P$, you have
$$
P(P/I)=(P^2+I)/I=P/I.
$$
A: I don't know what means a "trivial" answer for you, but this is indeed trivial: $P/I$ is an $R$-module, and $P$ an ideal, so what could mean $P(P/I)$? 
Well, let's consider a bigger picture: if $M$ is an $R$-module, then $PM$ is by definition the submodule of $M$ consisting of elements of the form $\sum a_im_i$ with $a_i\in P$ and $m_i\in M$. 
Coming back to our case, the elements in $P(P/I)$ are of the form $\sum a_i\bar x_i$ with $a_i\in P$ and $\bar x_i\in P/I$. But $\bar x_i\in P/I$ means that there is $p_i\in P$ such that $x_i-p_i\in I$, and therefore $\sum a_ix_i=\sum a_i(x_i-p_i)+\sum a_ip_i\in I+P^2$. Thus $\sum a_ix_i\in P$, hence $\sum a_i\bar x_i\in P/I$.
