The following solution is from an old blog entry of mine. Since $10\mid RHS$ we must have WLOG $(p_1,p_2)=(2,5)$. Plugging in and dividing by 10 yields $\prod_{i=1}^nq_i=7+\sum_{i=1}^nq_i$ for $q_i\in \bf P$. Clearly $n=1$ cannot work, so we try $n=2$ to get $q_1q_2=7+q_1+q_2\Leftrightarrow q_1q_2-q_2-q_2=7\Leftrightarrow (q_1-1)(q_2-1)=8=2^3$, and trying possible factors of 8 we find the unique solution $(q_1,q_2)=(3,5)$, or $(p_1,p_2,p_3,p_4)=(2,3,5,5)$. Now let $q_1\le q_2\le \cdots \le q_n$. Note that $\prod_{i=1}^nx_i\ge \sum_{i=1}^n x_i$ for any sequence of $\{x_i\}_{i=1}^n$ such that each $x_i\ge 2$. This follows from the fact that $x_1x_2\ge x_1+x_2\Leftrightarrow (x_2-1)(x_2-1)-1\ge 0$, and applying this fact repeatedly easily yields the result for higher $n$. Applying this to our $q_i$ equation, we see that $7+\sum_{i=1}^nq_i=\prod_{i=1}^nq_i\ge \left(\sum_{i=1}^{n-1}q_i\right)q_n$. Letting $P=\prod_{i=1}^{n-1}q_i$ and $S=\sum_{i=1}^{n-1}q_i$ we get $7+S+q_n\ge Sq_n\Leftrightarrow (q_n-1)(S-1)\le 8$. Testing the (highly limited) possible values of $q_n$ while keeping in mind that $n\ge 3\Leftrightarrow S\ge 2+2+2=8$ (since we already did the case $n=2$ above), we see that $(q_n-1)(S-1)\ge 7(q_n-1)\ge 7(1)=7<8$, however this assumes that $q_1=\cdots =q_4=2$ which (checking by hand) cannot happen, so $q_n\ge 3$ and $7(q_n-1)\ge 7(3-2)=7(2)=14>8$, contradiction. Thus, the only solution is the one we found above, namely $\boxed{(p_1,p_2,p_3,p_4)=(2,3,5,5)}$.

This is correct and very clear