# Exponentiation of mods

Could somebody at least tell me about mods in exponents?

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Mentallic
Homework Helper
Let 3^n denote 333... (n+1 threes in total)
Tetration already has a formalized expression.

$$^n3=3^{3^{...^3}}$$ with n 3's in the stack.

Or you can even use Knuth's up arrow notation. Exponentiation is denoted by 1 up arrow, and tetration by 2, so

$$^n3 = 3\uparrow\uparrow n$$

Tetration already has a formalized expression.

$$^n3=3^{3^{...^3}}$$ with n 3's in the stack.

Or you can even use Knuth's up arrow notation. Exponentiation is denoted by 1 up arrow, and tetration by 2, so

$$^n3 = 3\uparrow\uparrow n$$

I could only guess what you're asking for, but if I were to guess, say you want to prove that $^n3 \mod 10 = 7, n>1$ we would do an inductive proof. I'll make it short and you can fill in the details, but essentially you start with the base case $3^3 = 27$ hence $27 \mod 10 = 7$ and we can then show by calculation that $3^7 \mod 10 = 7$ hence it must be true for all n.