Mathematically speaking, the question is to find all primes $p_i$ such that $$p_1p_2p_3\cdots p_k = 1+p_1+p_2+\cdots p_k$$ for some positive integer $k$. I know that the only solutions are $2$ and $3$, but I am unable to figure out a well-written solution. Regarding the source of the question, I encountered it while proving that $6$ is the only square free perfect number. My proof for that went like this, assume $n$ is any perfect square free number and we wish to prove that $n=6$. Since, $n$ is square free, we can write the prime factorisation of $n$ as $$n=p_1p_2\cdots p_k$$ where all $p_i$ are distinct. Now, since $n$ is perfect, $$n=1+p_1+p_2+\cdots p_k$$ from which we get the original equation.
(PS: This is my first question on Stack Exchange so pardon my mistakes if any.)
As Qiaochu Yuan and Ross Millikan pointed out, I interpreted the meaning of a perfect number in a wrong way, so the original equation becomes $$2p_1p_2\cdots p_k = (1+p_1)(1+p_2)\cdots (1+p_k)$$ So, it will be helpful if someone brings out any solution to this or even the original problem of 6. Anyways, I am still interested in solutions to the original equation.
WLOG assume that $p_1\le p_2\le \dots\le p_k$. Then $p_1\cdots p_k\le 1+kp_k$. Since each $p_i\ge 2$, we have $2^{k-1}p_k\le 1+kp_k$ or $2^{k-1}\le 1/p_k+k\le k+1$. This implies $k=1$ or $2$. $k=1$ is clearly impossible. So $k=2$, $p_1p_2=1+p_1+p_2$. If $p_1=2$ we have $2p_2=3+p_2$, $p_2=3$.
Suppose that $p_1\ge 3$. So $3p_2\le 1+p_1+p_2$, $2p_2\le p_1+1$ which is impossible for primes $p_1\le p_2$. Hence the answer is $k=2, p_1=2, p_2=3$.