$$\boxed{\text{Define } n!! \overset{\text {def}}{=} n^{{(n-1)}^{{(n-2)}^{{{{\text{.}}^{\text{.}}}^1}}}} \text { for } n\in \mathbb{N}}$$
Find, with proof, all possible units digit of $n!!$ for $n\geq 1$
Pattern hunting (the unit's digit) or considering the repeat cycle of the unit's digit doesn't seem feasible in this case. Also, another way is to check $n$ modulo $5$ (and considering the parity of $n$) but that's quite strenuous and doesn't look good.
Am I missing something? Any hint or solution would be appreciated.
I think you are overestimating the pattern checking required. For example, if $n\equiv 0 \pmod {10}, n!! \equiv 0 \pmod {10}$ and we are done. If $n \equiv 9 \pmod {10}$ the exponent is even and $n!! \equiv 1 \pmod {10}$. The powers of each digit go through a cycle of at most $4$ when considered $\bmod 10$. For $2$ the power is always odd, so the only possibilities are $2,8$. We can see $2!!=2,$ so $2$ is possible. It is possible that the exponent is $3 \bmod 4$ when the last digit of $n$ is $2?$. Just do it and see where you run into problems.