It is easy to show that $2,5\mid a^5-a$, so $a^5\equiv a\pmod{10}$. It follows that $a^{x+4}\equiv a^x\pmod{10}$ for all integers $a,x\geq 1$. Next we show that $(n+20)^{n+20}\equiv n^n\pmod{4}$. This is clearly true when $n$ is even. When $n$ is odd, $n^2\equiv 1\pmod{4}$, so it is also true. Thus $(n+20)^{(n+20)^{n+20}}\equiv n^{n^n}\pmod{10}$, so it is periodic with period 20. Now we show 20 is the smallest period. Since $a_{10}=0$ and the next $a_i=0$ is $i=20$, so the smallest period is either 10 or 20. Note that $3^{3^3}\equiv 3^3\equiv 7\pmod{10}$ but $13^{13^{13}}\equiv 3^1\equiv 3\pmod{10}$, so its period is not 10. Thus the smallest period is 20.