HINT: Suppose that you have a four-digit number $n$ that is written $abcd$. Then
$$\begin{align*}
n&=10^3a+10^2b+10c+d\\
&=(999+1)a+(99+1)b+(9+1)c+d\\
&=(999a+99b+9c)+(a+b+c+d)\\
&=3(333a+33b+3c)+(a+b+c+d)\;,
\end{align*}$$
so when you divide $n$ by $3$, you’ll get
$$333a+33b+3c+\frac{a+b+c+d}3\;.$$
The remainder is clearly going to come from the division $\frac{a+b+c+d}3$, since $333a+33b+3c$ is an integer.
Now generalize: make a similar argument for any number of digits, not just four. (If you know about congruences and modular arithmetic, you can do it very compactly.)
I believe your question is still open. Some notation before we move on to a related conjecture:
Denote $a\preceq b$ if $a$ is a subsequence of $b$ when they are both written in base $10$, e.g., $553\preceq 65536$, $63\preceq 65536$, but $35$ is not a subsequence (so order matters). Given a set $S\subseteq\mathbb{N}$, we define the minimal set as the smallest set $M\subset S$, such that for all $s\in S$, there exists $m\in M$ and $m\preceq s$. From Higman's lemma, we know that the minimal set is finite for every possible $S$.
Back to your question, there is a weaker conjecture by J. Shallit in
Minimal Primes, J. Recreational Math. 30 (2000), 113-117
that $\{1,2,4,8,65536\}$ is the minmal set of powers of $2$ in base $10$, which as far as I can tell is still open. As Shallit observed in the paper that this conjecture is true if "every power of $16$ greater than $65536$ contains a digit in the set $\{1,2,4,8\}$", which is precisely your question.
For numerical search, you can take a look at $A137998$. The question is equivalent to the statement that $16^4$ is the only entry of $0$ that appears in this sequence. Due to a repeating pattern mentioned in its comments, you would only need to check at most $12$ out of every $25$ powers of $16$.
(Note this had been posted as a comment, but as it was the 10+th comment, I know that not everyone would notice it. I hope it has been expanded enough to be warranted as an answer. Please let me know if otherwise)
edited: below is taken from Shallit's paper as an attempt to make the concept of "minimal set" more accessible.
We will take a look at prime numbers. If I were to claim, as an analogy to the original question, that all prime numbers when written in base-$10$ must contain a prime digit, i.e., $\{2,3,5,7\}$, you would easily be able to find a counterexample: "How about $11$? and $19$?". However, I could expand the list to include all $2$-digit prime numbers that has no prime digit, i.e., $\{2,3,5,7,11,19,41,61,89\}$. Then I would claim that all prime numbers must contain a prime number in this new list. This is where I would need to clarify what it means by a number to "contain" another number with more than one digit, e.g., $19$. The defintion chosen here is that given a number $b$, if we can reach $a$ by deleting some digits of $b$, then we say $b$ contains $a$ (or $a\preceq b$). As an example, $109$ is a prime number and it contains $19$. This time it would take you a bit effort to find a disproof: the first one is $409$. The procedure goes on as I would extend the list to include such $3$-digit prime numbers and you would produce larger counter-examples, so on so forth.
The surprise is that this process necessarily ends, due to Higman's lemma. As my list grows longer (but remains finite), at certain point the claim becomes true. We call the final list minimal set of prime numbers.
We can repeat this process for powers of $2$, which leads to the conjecture that $\{1,2,4,8,65536\}$ is the minimal set. This is apparently weaker than your question, as it allows the possibility that there exists some power of $2$ that does not contain $1,2,4,8$ but contains $65536$. Note the proofs we know for Higman's lemma, albeit constructive, are not effective to give us the actual minimal set in general. Shallit did manage to get the answer for prime numbers using elementary means, but as far as I know, there is no progress for the minimal set of powers of $2$, which in turn indicates that the stronger statement in your question is still open.
Best Answer
Not quite right, since $9\times 0 = 0$ and the digits don't add up to $9$; but otherwise correct.
The reason it works is that we write numbers in base $10$, and when you divide $10$ by $9$, the remainder is $1$. Take a number, say, $184631$ (I just made it up). Remember what that really means: $$184631 = 1 + 3\times 10 + 6\times 10^2 + 4\times 10^3 + 8\times 10^4 + 1\times 10^5.$$ The remainder when you divide any power of $10$ by $9$ is again just $1$, so adding the digits gives you a number that has the same remainder when dividing by $9$ as the original number does. Keep doing it until you get down to a single digit and you get the remainder of the original number when you divide by $9$, except that you get $9$ instead of $0$ if the number is a nonzero multiple of $9$.
But since every multiple of $9$ is a multiple of $9$, you will always get $9$.
Note that you have a similar phenomenon with $3$ (a divisor of $9$), since adding the digits of a multiple of $3$ will always result in one of the one-digit multiples of $3$: $3$, $6$, or $9$.
If we wrote in base $8$, instead of base $10$, then $7$ would have the property: if you write a number in base $8$ and add the digits (in base 8) until you get down to a single digit between $1$ and $7$, then the multiples of $7$ will always end yielding $7$, for precisely the same reason. And if we wrote in base $16$, then $15$ (or rather, F) would have the property. In general, if you write in base $b$, then $b-1$ has the property.
This is a special case of casting out nines, which in turn is a special case of modular arithmetic. It is what is behind many divisibility tests (e.g., for $2$, $3$, $5$, $9$, and $11$).
Coda. This reminds me of an anecdote a professor of mine used to relate: a student once came to him telling him he had discovered a very easy way to test divisibility of any number $N$ by any number $b$: write $N$ in base $b$, and see if the last digit is $0$. I guess, equivalently, you could write $N$ in base $b+1$, and add the digits to see if you get $b$ at the end.