Finding the smallest divisor Find the smallest divisor of $ 12!+6!+12!\cdots6!+1!$ except 1 
I know this has to do something with Wilson's theorem which states the if $n$ is a prime number $n$ will divide $(n-1)!+1$.

It is straightforward to check that 1 to 6 are no divisors of the expression, since they are factors of $12!$, $6!$ and $12!6!$.
Further, 7 is a divisior of both $12!$ and $12!6!$. Because $$12!=1\cdot2\cdot3\ldots6\cdot\color{blue}{7}\cdot8\ldots 11\cdot12$$
By Wilson's theorem, we have $$ 6! \equiv -1 \mod 7$$
so $6!+1$ is also divisible by 7. Therefore $7 \mid 12!6!+12!+6!+1$
