since $\gcd(7,9 = 1)$, so we have two cases:
$d = 7$, trivially divisible by $7$
$10^{n!} - 1 \equiv 0 \pmod{7} \implies 10^{n!} \equiv 1 \pmod{7}$
We note that $6$ is the order of $10 \in \mathbb{Z}_7$, by Fermat's Little Theorem (or Euler's Totient Theorem), we must have $6 \mid n! \implies n \geqslant 3$
since $\gcd(7,9 = 1)$, so we have two cases:
$d = 7$, trivially divisible by $7$
$10^{n!} - 1 \equiv 0 \pmod{7} \implies 10^{n!} \equiv 1 \pmod{7}$
We note that $6$ is the order of $10 \in \mathbb{Z}_7$, by Fermat's Little Theorem (or Euler's Totient Theorem), we must have $6 \mid n! \implies n \geqslant 3$