<s>[INST] Determine the last digit of the following number. Problem: Determine the last digit of the following number $$\underbrace{7^{7^{7^{7...}}}}_{1001\text{ }7's}.$$
My Attempt: By Euler's theorem $7^4\equiv 1\pmod {10}\Rightarrow 7^7\equiv 3\pmod {10}.$ After this I am clueless.
Edit $1$: So I tried taking even and odd number of sevens and this is what I observed $$7^{7}\equiv 3\pmod{10}.$$ $$7^{7^{7}}\equiv 7\pmod{10}.$$ $$7^{7^{7^{7}}}\equiv 7^{7^{7*7^6}}\equiv \left(7^{7^{7}}\right)^{7^6}\equiv 7^{7^6}\equiv 3\pmod{10}.$$ Thus for an odd number of sevens it seems that the remaineder will be $7$ whereas for an even number of sevens the remainder will be $3.$
Edit $2$: We need to find $7^7\equiv 1\pmod 4$ therefore $7^{7^{7}}\equiv 7^{4k+1}\equiv 7\pmod {10}.$ Similarily $7^{7^{7^{7}}}\equiv 7^{7^{4k+1}}\equiv 7^7\equiv 3\pmod {10}.$ Also observe that $7^{7^{7}}\equiv 7^{2k+1}\equiv 3\pmod 4.$ If we have $5$ sevens $7^{7^{7^{7^7}}}\equiv 7^3\equiv 3\pmod{10}.$
 [/INST] As you observed, by Euler's theorem, the exponent works modulo $4$.  And $7$ happens to be congruent to $-1 (\bmod{4})$.  The exponent is a tower of $1000$ $7$'s, which, modulo $4$, is $-1$ raised to an odd power.  So modulo $10$, your tower of $1001$ $7's$ is $\equiv 7^{-1} \equiv 3 (\bmod{10}).$
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