Calculating 7^7^7^7^7^7^7 mod 100 What is
$$\large 7^{7^{7^{7^{7^{7^7}}}}} \pmod{100}$$
I'm not much of a number theorist and I saw this mentioned on the internet somewhere. Should be doable by hand.
 A: Reading the other answers, I realize this is a longer way than necessary, but it gives a more general approach for when things are not as convenient as $7^4\equiv 1\bmod 100$. 
Note that, for any integer $a$ that is relatively prime to $100$, we have
$$a^{40}\equiv 1\bmod 100$$
because $\varphi(100)=40$, and consequently
$$a^m\equiv a^n\bmod 100$$
whenever $m\equiv n\bmod 40$. Thus, we need to find $7^{7^{7^{7^{7^{7}}}}}$ modulo $40$. By the Chinese remainder theorem, it is equivalent to know what it is modulo $8$ and modulo $5$.
Modulo $8$, we have $7\equiv -1\bmod 8$, and $-1$ to an odd power is going to be $-1$, so we see that $$7^{7^{7^{7^{7^{7}}}}}\equiv (-1)^{7^{7^{7^{7^{7}}}}} \equiv -1\equiv 7\bmod 8.$$
Modulo $5$, we have $7^4\equiv 1\bmod 5$ (again by Euler's theorem), so we need to know $7^{7^{7^{7^{7}}}}\bmod 4$. But $7\equiv -1\bmod 4$, and $7^{7^{7^{7}}}$ is odd, so that $7^{7^{7^{7^{7}}}}\equiv -1\equiv 3\bmod 4$, so that 
$$7^{7^{7^{7^{7^{7}}}}}\equiv 7^3\equiv 343\equiv 3\bmod 5.$$
Applying the Chinese remainder theorem, we conclude that
$$7^{7^{7^{7^{7^{7}}}}}\equiv 23\bmod 40,$$
and hence
$$7^{7^{7^{7^{7^{7^{7}}}}}}\equiv 7^{23}\bmod 100.$$
This is tractable by again using the Chinese remainder theorem to find
$7^{23}\bmod 4$ and $7^{23}\bmod 25$.
A: $7^4 = 2401 \equiv 1 \pmod{100}$, so you only need to calculate $7^{7^{7^{7^{7^7}}}} \pmod{4}$. We know that $7 \equiv -1 \pmod 4$ and $7^{7^{7^{7^7}}}$ is odd, so $7^{7^{7^{7^{7^7}}}} \equiv -1 \equiv 3 \pmod 4$, and then $$7^{7^{7^{7^{7^{7^7}}}}} \equiv 7^3 \equiv 43 \pmod {100}$$
A: A quick hand calculation gives
$$\begin{align}
7^1 &\equiv 7 \pmod{100} \\
7^2 &\equiv 49 \pmod{100} \\
7^3 &\equiv 43 \pmod{100} \\
7^4 &\equiv 1 \pmod{100}
\end{align}$$
So it reduces to the problem of calculating the value of $7^{7^{7^{7^{7^7}}}} \pmod 4$. And $7^2 \equiv 1 \pmod 4$, so it reduces to the problem of calculating $7^{7^{7^{7^7}}} \pmod 2$... and this is easy, it's odd, so it's congruent to $1$ modulo $2$.
Working backwards:
$$7^{7^{7^{7^{7^7}}}} \equiv 7^1 \equiv 3 \pmod{4}\quad \Rightarrow\quad 7^{7^{7^{7^{7^{7^7}}}}} \equiv 7^3 \equiv 43 \pmod{100}$$
