Arrow Notation

This problem can actually be solved with just logs!

$3\uparrow \uparrow 3={ 3 }^{ { 3 }^{ 3 } }={ 3 }^{ 27 }$

Evaluating modulo $10$ , we get that the last digit is $7$ . This is because $27 = 3 \pmod 4$ , and ${3}^{n}$ repeats in cycles of $4$ modulo 10.

Now you need to find the first digit to identify which of the remaining options could work. I used that $\log _{ 10 }{ 3 } \approx 0.477121$ .

Thus, $\log _{ 10 }{ {3}^{27} } \approx (0.477121)(27) = 12.882...$ . From there, the first digit must be $\lfloor{10}^{0.882} \rfloor = 7$ . The only remaining option is $7625597484987$ .

$3\uparrow \uparrow 3={ 3 }^{ { 3 }^{ 3 } }={ 3 }^{ 27 }=7625597484987$

See Knuth's up-arrow notation .