Find the last decimal digit of \[ 7^{\,7^{\,7^7}}. \]
Solution
Let \[ E\coloneqq7^{7^7}. \]
1. One has\[ E\equiv3\pmod4. \]
Since \[ 7\equiv3\pmod4 \] and \(7^7\) is odd, \[ E = 7^{7^7} \equiv 3^{7^7} \equiv 3 \pmod4. \]
2. Powers of \(7\) modulo \(10\) have period \(4\).
Directly, \[ 7^1\equiv7, \qquad 7^2\equiv9, \qquad 7^3\equiv3, \qquad 7^4\equiv1 \pmod{10}. \] Multiplication by \(7^4\equiv1\) repeats the cycle.
3. The last decimal digit of the given power tower is\[ \boxed{3}. \]
By step 1, \[ E\equiv3\pmod4. \] By the period in step 2, \[ 7^E \equiv 7^3 \equiv 3 \pmod{10}. \] Thus the last decimal digit is \(3\).
4. Q.E.D.
Step 3 gives the requested digit.