Let the highest power of prime $p$ in $a,b,c$ be $A,B,C$ respectively.
As $ab$ is perfect square, $A+B$ is even
Similarly, $B+C$ will be even $\implies (A+B)+ (B+C)$ will be even
Now the highest power of $p$ in $ac$ will be $A+C$
Now as $A+C-(A+B+B+C)=2B$ is even, so will be $A+C$
A comment on strategy too long for the comment section, but not a complete answer:
You want to find $6!\prod a_i! +1=x^2$ where $a_i$ are single digits. Note that the single digits are $0,1,2,3,2^2,5,2\cdot 3,7,2^3,3^2$, so the factorials will all either be $1$ or will have only the prime factors $2,3,5,7$.
Rearranging, $6!\prod a_i!=x^2-1=(x-1)(x+1)$. LHS is even, so $(x-1)\ \text{and}\ (x+1)$ are both even. Successive even numbers have only a single factor of $2$ in common.
$2^2\cdot3^2\cdot5\cdot\prod a_i!=\frac{(x-1)}{2}\frac{(x+1)}{2}$. RHS are two consecutive numbers; hence their gcd is $1$. So the first problem is to find consecutive numbers which only have prime factors of $2,3,5,7$, and none in common.
$161$ leads to a solution because $\frac{161-1}{2}=80=2^4\cdot 5;\ \frac{161+1}{2}=81=3^4$.
$17$ leads to a solution (although not for the $6$ question) because $\frac{17-1}{2}=8=2^3;\ \frac{17+1}{2}=9=3^2$.
Finding candidate consecutive numbers will not necessarily lead to a solution, as the numbers of various prime factors will have to be distributable among the $a_i!$, which might prove difficult if (for example) the number of factors of $2$ is less than the number of factors of $3$ in the consecutive numbers. If consecutive numbers can be found that satisfy that requirement, then identifying suitable $a_i$ will be possible, and as OP points out, it should then be possible to generate a prime starting number by strategic insertion of digits $0,1$.
Best Answer
Let $a=9+4\sqrt{5}$, then $$f(n) = {5\over 64}(a^n-a^{-n})^2$$
Now let $$b_n = {\sqrt{5}\over 8}(a^n-a^{-n})$$
so it is enought to prove that every $b_n$ is an integer. This can be done easly if you write a recursive formula for $b_n$:
$$b_{n+1}= 18b_n-b_{n-1}$$ where $b_0=0$ and $b_1=5$ and prove that fact with induction.