Finding the last digit of $103^{103^{103^{103^{103}}}}$ I need to find the last digit of $103^{103^{103^{103^{103}}}}$ so the value in $\mod10$. I know
\begin{align}
103^{103^{103^{103^{103}}}}&=(100+3)^{103^{103^{103^{103}}}}\\
&=100\cdot(stuff)+3^{103^{103^{103^{103}}}}\\
&=3^{103^{103^{103^{103}}}} \mod10
\end{align}
And then I am stuck. I have observed that the last digit of a power of 3 is cyclic:
$$3 \to 9 \to 7 \to 1 \to 3 \to 9 \to 7 \to\cdots$$
This means it depends on $103^{103^{103^{103}}}\mod4$.
Is this the right method? do I just need to keep doing this or is there a simpler "trick"?
 A: Hint: $103=-1\pmod4$, and $-1$ to any odd power is $-1$.
Edit: Full answer: once you arrive at $3^{103^{103^{103^{103}}}}$, you can take the exponent mod 4, like you suggested. Drawing from my hint, we have $103^{103^\text{positive int}}=(-1)^{\text{odd number}}=-1=3\pmod 4$, since the exponent $103^\text{positive int}$ is an odd number. So your answer is $3^3=7\pmod{10}$. The last digit is $7$.
A: You're on the right track.
Your observed pattern can be explained by $3^4=81\equiv1\pmod{10}$.

$103^{103}\equiv3^3\equiv3\pmod{4}$, so $103^{103^{103}} \equiv103^3\equiv3^3\equiv7\pmod{10}$.
$103^{103^{103}}\equiv3^3\equiv3\pmod4,$ so $103^{103^{103^{103}}}\equiv103^3\equiv3^3\equiv7\pmod{10}.$
Can you observe another pattern now?
A: $\color{#0a0}{k\ \rm odd},\ (a,10)\!=\!1\overset{\rm Euler}\Rightarrow\bmod 10\!:\ a^{\large\color{#c00} 4}\equiv 1\,\overset{\rm\color{#90f}{MOR}}\Rightarrow\, a^{\large (4n-1)^{\Large k}}\!\!\equiv a^{\large (\overbrace{4n-1)^{\Large k}\bmod\color{#c00} 4}^{\!\!\!\!\!\Large \equiv\ \,\color{#0a0}{(-1)^{k}\  \equiv}\,\ \LARGE\color{#0a0}3 }}\!\equiv a^{\large \color{#0a0}3}$
by Euler's $\phi$ Theorem combined with $\,\rm\color{#90f}{MOR}= \,$ Modular Order Reduction.
The OP is the special case $\, a = 103 = 4n\!-\!1,\,\ k = 103^{103^{103}}\! $ (or any odd integer)
