This is a special case of casting out nines, which is best viewed in terms of modular arithmetic:
$\rm\quad mod\ 9:\ \ 10\equiv 1\ \Rightarrow\ P(10)\equiv P(1)\quad $ for all polynomials $\rm\,P(x)\,$ with integer coefficients
by the Congruence Polynomial Rule.
But radix notation has polynomial form, e.g. $\rm\ 567 = P(10),\ \ P(x) = 5\: x^2 + 6\: x + 7.\, $ Therefore, the above implies that, mod $9$, an integer $\rm\ N = P(10)\equiv P(1) = $ the sum $\rm\,S_N\,$ of its decimal digits.
Yours is the special case $\rm\ N\equiv 0\ \ (mod\ 9)\ $ so the above implies that it remains $\equiv 0\ $ when you map $\rm\:N\:$ to its digit sum $\rm\:S_N.\:$ Since this map is strictly decreasing for $\rm\:N > 9,\ $ iterating it must eventually reach some nonzero $\rm\: N' \le 9.\ $ But $\rm\ N'\equiv 0\ \ (mod\ 9)\ $ so we conclude that $\rm\: N' = 9.$
Note: if modular arithmetic is unfamiliar then one may proceed more simply as follows:
Factor Theorem $\rm\;\Rightarrow\; X-1\ |\ P(X)-P(1)\ $ so $\rm\ X = 9\ \Rightarrow\ 9\ |\ P(10)-P(1) \;\:$ i.e. $\rm\; 9\ |\ N - S_N$.
The analogous result holds true for any radix $\rm\:b,\,$ i.e. one can cast $\rm\:b-1$'s in the same way, since $\rm\ P(b) \equiv P(1)\ \ (mod\ b-1)\:.\:$ Therefore $\:9\:$ is "special" only in the way it relates to the radix $10\:$. Similarly we may cast $11$'s using $\rm\ P(10)\equiv P(-1)\ \ (mod\ 11)\:.\:$ One can devise countless divisibility tests by using modular reduction in this way, e.g. see here for casting out $91$'s.
It deserves to be better known that one may also cast out nines to check rational arithmetic - as long as the fractions have denominator coprime to $3$, e.g. see Hilton; Pedersen: Casting out nines revisited. Analogous remarks hold true for any ring that has $\ \mathbb Z/9\ $ as an image - just as one can apply parity arguments in any ring that has $\ \mathbb Z/2\ $ as an image, e.g. all rationals with odd denominator, or the Gaussian integers $\rm\:\mathbb Z[i]\:$, where the image $\rm\ \mathbb Z[i]/(2,i-1) \cong \mathbb Z/2\ $ yields the natural parity definition that $\rm\ a+b\:i\ $ is even iff $\rm\ a\equiv b\ \ (mod\ 2)\ $, i.e. if $\rm\ a+b\:i\ $ maps to $\:0\:$ via the above isomorphism, which maps $\rm\ 2\to 0,\ i\to 1\:$. See here for further discussion of parity in rings of algebraic integers, including examples of number rings with no parity structure, and with more than one parity structure. See also this post for "casting out orders" in cyclic groups, and see this thread for an in-depth comparison of various elementary inductive proofs of casting nines.
These are elementary prototypical examples of problem-solving by way of modular reduction - one of the keystones of abstract algebra. As such one should be sure to understand these simple instances before moving on to more advanced manifestations of modular reduction.