Lemma 1
$$ \forall n \in \mathbb{N}_{>0} \hspace{1.5em} n^{n^n} \equiv n^n \pmod3 $$
This is trivial.
Lemma 2
$$\forall n \in \mathbb{N} \hspace{1.5em} n\ge 3 \implies n^{n^n} \equiv n^n \pmod{16} $$
Proof. If $n$ is even, then $n\ge4$ and thus $16 \mid n^n, n^{n^n}$. If $n$ is congruent to $\pm 1$ modulo $16$, the proposition above holds. Then, we only need to check $\pm 3, \pm 5, \pm 7$. By Euler's theorem,
\begin{align}
n\equiv 3 \pmod{16} \implies 3^{n^n} - 3^n &\equiv 3^{n^n-n \pmod8}-1 \\
&\equiv 3^{3^{n\pmod4}-3}-1 \\
&\equiv 3^{3\cdot 8} -1 \equiv 0 \pmod{16}
\end{align}
\begin{align}
n\equiv 5 \pmod{16} \implies 5^{n^n} - 5^n &\equiv 5^{n^n-n \pmod8}-1 \\
&\equiv 5^{5^{n\pmod4}-5}-1 \equiv 0 \pmod{16}
\end{align}
\begin{align}
n\equiv 7 \pmod{16} \implies 7^{n^n} - 7^n &\equiv 7^{n^n-n \pmod8}-1 \\
&\equiv 7^{7^{n\pmod4}+1}-1 \\
&\equiv 7^{7^3+1} -1 \equiv 7^{43\cdot 8} -1 \equiv 0 \pmod{16}
\end{align}
The cases $-3,-5,-7$ are now trivial.
Theorem
$$\forall n \in \mathbb{N} \hspace{1.5em} n\ge 3 \implies n^{n^{n^n}} \equiv n^{n^n} \pmod{1989} $$
Proof. We first split the congruence as follows.
$$\begin{align} n^{n^{n^n}} \equiv n^{n^n} \pmod{1989} &\iff \begin{cases}
n^{n^{n^n}} \equiv n^{n^n} \pmod{9} \\
n^{n^{n^n}} \equiv n^{n^n} \pmod{13} \\
n^{n^{n^n}} \equiv n^{n^n} \pmod{17}
\end{cases}
\end{align}$$
Since $2,2,3$ are primitive roots of $9,13,17$ respectively, then
$$\begin{align}
n^{n^{n^n}} \equiv n^{n^n} \pmod{9} &\iff \begin{cases} n^{n^n}\cdot\operatorname{ind}_2n \equiv n^n\cdot\operatorname{ind}_2n \pmod{6} & 3 \nmid n \\ 0 \equiv 0 & 3 \mid n\end{cases} \\[1ex]
n^{n^{n^n}} \equiv n^{n^n} \pmod{13} &\iff \begin{cases} n^{n^n}\cdot \operatorname{ind}_2n \equiv n^n \cdot\operatorname{ind}_2n \pmod{12} & 13 \nmid n \\ 0 \equiv 0 & 13 \mid n\end{cases} \\[1ex]
n^{n^{n^n}} \equiv n^{n^n} \pmod{17} &\iff \begin{cases} n^{n^n}\cdot \operatorname{ind}_3n \equiv n^n \cdot\operatorname{ind}_3n \pmod{16} & 17 \nmid n \\ 0 \equiv 0 & 17 \mid n\end{cases}
\end{align}$$
By the lemmas, these congruences hold for all $n\ge 3$.