We consider $n^{n^{n^n}}=n^{n^n}$ in $\mathbb{Z}_m$ for coprime integers m=9, 13, 17, and the result will follow from the Chinese Remainder Theorem. Lemma: If $n$ is coprime with $m$, then, dividing by $n^{n^n}$ (which I may do since m and n are coprime), I must prove that $n^{n^{n^n}-n^n}\equiv 1\mod m$, which one can reduce, by Euler-Fermat, to showing that $\phi(m)|n^{n^n}-n^n$ For 9, if 3|n it is trivial, so we consider when 3 and n are coprime. Using the lemma, we must show $6|n^{n^n}-n^n$. Modulo 2 this is either 0-0 or 1-1, both of which are clearly zero. Modulo 3, since $n^2\equiv 1$ by FLT, if $n$ is even we get 1-1, and if $n$ is odd we get n-n, so in both cases we get a zero, so $2*3|n^{n^n}-n^n$ and we are done. For 13, again, if 13|n then it is trivial, so we assume 13 and n are coprime. Again we can apply the lemma and must show $12|n^{n^n}-n^n$, but we showed 3 works already, so it remains to show $4|n^{n^n}-n^n$. If n is even this is obvious. If n is odd, then $n^2 \equiv 1 \mod 4$ so $n^{n^n}-n^n \equiv n-n \equiv 0$ and we are done. For 17, clearly if 17 divides n then we are done, so let us suppose it is not, so n must be coprime to 17. If $n$ is even, $n\geq 4$, so $16|n^n$, since $2^4=16$. But then by Fermat's Little Theorem the above equation reduces to $1=1$ which is clearly true. If $n$ is odd, then I use the lemma. $n^{n^n}-n^n=n^n(n^{n^n-n}-1)$. Now, since $n^2\equiv 1 \mod 8$ for all odd n, $n^n\equiv n \mod 8$ for all odd n, so $8|n^n-n$, and $\phi(16)=8$, so since 2|n, n being coprime to 16, Euler's Theorem states that $(n^{n^n-n}-1) \equiv 0 \mod 16$, which is $\phi(17)$, so we are done.

$ 1989 = 3^2 \cdot 13 \cdot 17$, hence $ 1989 \mid n^{n^{n^n}} - n^{n^n} \iff 3^2,13,17 \mid n^{n^{n^n}} - n^{n^n}$ First $ n^{n^{n^n}} \equiv n^{n^n} \bmod 3^2$. If $ 3 \mid n$ we are done, so wlog assume $ (3,n) = 1$. By euler-fermat it is enough to prove that $ n^{n^n} \equiv n^n \bmod \varphi(3^2) = 6$. That is: $ n^{n^n} \equiv n^n \bmod 2 (*)$ and $ n^{n^n} \equiv n^n \bmod 3 (**)$ $ (*)$ is obvious since it suffices to consider $ n$ odd and even. For proving $ (**)$ we again note that $ 3 \mid n$ makes it obvious, so it is enough to prove $ n^n \equiv n \bmod \varphi(3) = 2$. And $ n^n \equiv n \bmod 2 (***)$ is obvious by parity too. $ n^{n^{n^n}} \equiv n^{n^n} \bmod 13$. It is again enoughh to prove $ n^{n^n} \equiv n^n \bmod \varphi(13) = 12$ that is: $ n^{n^n} \equiv n^n \bmod 4 (****)$ and $ n^{n^n} \equiv n^n \bmod 3 (**)$. We proved $ (**)$ beforehand and $ (****)$ follows by considering $ 2 \mid n, n \equiv \pm 1 \bmod 4$. $ n^{n^{n^n}} \equiv n^{n^n} \bmod 17$. It is enough to prove $ n^{n^n} \equiv n^n \bmod \varphi(17) = 16$. From there it is enough to prove $ n^n \equiv n \bmod 8$ when $ 2 \nmid n$. This follows again by parity, and the fact that $ x^2 \equiv_8 1$ when $ (x,2) = 1$.

Note that $1989=3^2 \cdot 13\cdot 17$. Checking $9$: For $n\equiv 3,6 \pmod 9$ we obviously have $n^{n^{n^n}}\equiv n^{n^n} \pmod 9$. In the other cases just use Euler's totient theorem, noting that $\phi(9)=6$. Checking $13$: It suffices to prove $n^{n^n}\equiv n^n \pmod{\phi(13)}$. We have $\phi(13)=12$. It's obvious when $n$ is even or divisible by $3$. So we need to check $n\equiv 5,7,11\pmod 12$. Just use Euler's totient again. Checking $17$: Pretty similar to previous case.