Proving $19 \mid 2^{2^n} + 3^{2^n} + 5^{2^n}$ Theorem. $19 \mid 2^{2^n} + 3^{2^n} + 5^{2^n}$, for all positive integers $n$.

I'm tasked with proving the given theorem by induction. Here's where I've gotten so far...
Proof. Clearly, the theorem is true for $n=1$, establishing the base case. Moreover, it can easily be shown that the theorem works for $n=2,3,4,5,\ldots, 18$.
Suppose, now, that $k$ is an integer for which the given theorem holds. Consider, then, $k+18$.
$$2^{2^{k+18}} + 3^{2^{k+18}} + 5^{2^{k+18}}$$
By Fermat's Little Theorem, if $n \equiv m \pmod{p-1}$, then $a^m \equiv a^n \pmod{p}$. Clearly, $k+18 \equiv k \pmod{18}$. Then,
$$2^{2^{k+18}} \equiv 2^{2^k} \pmod{19}$$
$$3^{2^{k+18}} \equiv 3^{2^k} \pmod{19}$$
$$5^{2^{k+18}} \equiv 5^{2^k} \pmod{19}$$
Hence, $2^{2^{k+18}}$ may be expressed as $\;2^{2^k} + 19n$ for some integer $n$, $\;3^{2^{k+18}}$ as $\;3^{2^k}+ 19m$ for some integer $m$, and $\;5^{2^{k+18}}$ as $\;5^{2^k} + 19q$ for some integer $q$.
Thus, $2^{2^{k+18}} + 3^{2^{k+18}} + 5^{2^{k+18}} = (2^{2^{k}} + 3^{2^{k}} + 5^{2^{k}}) + 19(m+n+q)$, a sum of integers divisible by $19$, and thus clearly divisible by $19$.
By the principle of induction, we've thus shown that $19 \mid 2^{2^n} + 3^{2^n} + 5^{2^n}$, for all positive integers $n$, concluding the proof.
$\blacksquare$

Is this approach valid?

I don't think it's valid.  $k \equiv k + 18 \pmod {18}$ so $2^k \equiv 2^{k+18} \pmod {19}$.  And that means $a^{2^k} \equiv a^{2^{k+18}} \pmod m$ for an $m$ where $\phi(m) = 19$ and $\gcd(a,m) =1$ but... that doesn't work for  $a^{2^{k+18}}\equiv a^{2^k} \pmod {19}$
But you could be onto something.
$2^{k}\pmod{18}$ cycles.  $\phi 9 = 6$ so $2^{k+ 6} \equiv 2^k \pmod 9$ and $2^m \equiv 0 \pmod 2$ so for $k \ge 1$ we know $2^{k+ 6}\equiv 2^k$ and so $a^{2^{k+6}}\equiv a^{2^{k}} \pmod {19}$.
So we can do $6$ base cases.  Easier than $18$.
For $k = 1, ....6$ we have $2^k \equiv 2, 4,8,16, 14, 10 \pmod {18}$ and
But we can do better.
$2^2\equiv 4; 2^4 \equiv 16\equiv -3; 2^8 \equiv 9 \equiv 3^2\pmod {19}$
$3^2 \equiv 9; 3^4 \equiv 81 \equiv 5; 2^8 \equiv 5^2\pmod {19}$.
And $5^2 \equiv 6; 5^4 \equiv 36 \equiv -2$ and $5^{8}\equiv 2^2\pmod {19}$.
SO for $k \ge 1$ we have
$2^{2^{k+2}} + 3^{2^{k+2}} + 5^{2^{5+2}} \equiv 3^{2^k} + 5^{2^k} + 2^{2^k}\equiv 2^{2^k} + 3^{2^k} + 5^{2^k}\pmod {19}$
So we just need 2 base cases:
$2^2 + 3^2 + 5^2 \equiv 4+9 + 6 \equiv 0 \pmod {19}$
$2^{2^2} + 3^{2^2} + 5^{2^2} \equiv 16 + 81 + 36 \equiv -3+5 -2\equiv 0 \pmod{19}$
That's it.
