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?

Note that incrementing $n$ just corresponds to squaring each summand.  Just compute the first few cases mod 19 and look for a pattern:

*

*For $n=1$, we have $2^2 \equiv 4$, $3^2\equiv 9$, $5^2 \equiv 6$.  We are good since $4+9+6\equiv 19$.

*For $n=2$, we have $2^3=4^2 \equiv -3$, $3^3=9^2 \equiv 5$, $5^3\equiv 6^2 \equiv -2$.  Good again since $-3+5-2 \equiv 0$.

Now note that squaring the summands of the $n=2$ case will just give us the $n=1$ case again, as the numbers are the same as what we started with (2, 3, and 5) up to signs.  So subsequent increments of $n$ will just cause the summands to alternate between {4, 9, 6} and {-3, 5, -2}.  So we're done.
