elementary number theory - Checking the remainder of $40^{1999}$ divided by $7$

I have been learning some applications to Fermat's Little theorem and I am currently solving the remainder of $40^{1999}$ divided by $7$ and would appreciate if anyone could confirm my answer of $5$.

$7$ is prime and $7\nmid 40$

So according to Fermat's Little Theorem

$40^{6} \equiv 1 \pmod{7}$

$1999/6 = 333\times 6 + 1 $

Therefore $(40^6)^{333}\times 40^1 \equiv 1^{333}\times 40 \pmod{7}$

$40^{1999} \equiv 40 \pmod{7}$

$40^{1999} \equiv 5 \pmod{7}$

The remainder of $40^{1999}$ divided by $7$ is $5$