why does $3^{10^n} \text{ mod } 10^{n+2} = 2 \cdot 10^{n+1}+1$? For integers n > 1 it seems that: $$3^{10^n} \text{ mod } 10^{n+2} = 2 \cdot 10^{n+1}+1$$
For example, the last 12 digits of $3^{10^{10}}$ are 200000000001.
I have verified that is true at least up to n = 250 and want a way to prove that it is true for all integers n > 1.
It also appears that many more of the last digits of $3^{10^n}$ are the same.
$3^{10^{10}} = \quad ...228442786552200000000001$
$3^{10^{20}} = \quad ...8084427865522000000000000000000001$
 A: HINT: Your observation about the ending digits is very useful. If you can prove that the string of digits $2000...0001$ contains a total of $n+2$ digits (including the $1$ at the end) then the number minus $2\cdot10^n+1$ would have $n+2$ zeroes at the end and would thus be divisible by $10^{n+2}$. So if you can prove that all numbers in the form that you mentioned end with a string of $n+2$ digits in the form
$$...2000...0001$$
then the rest of the proof comes easily.
A: You can try induction on $n$. 
$$3^{10^n} = k_n10^{n+2}+2\cdot 10^n+1$$
for some integer $k_n$.
It's true for $n=2$, by your calculation. Now assume true for some $n\geq 2$. Then applying the binomial theorem:
Then $$3^{10^{n+1}}=\left(3^{10^n}\right)^{10}=(k_n10^{n+2}+2\cdot 10^n +1)^{10}= C10^{n+3}+(2\cdot 10^n+1)^{10}$$
But $$(2\cdot 10^n+1)^{10}=1+\binom{10}{1}2\cdot 10^n+ \binom{10}{2}2^2\cdot (10^{2n})+\cdots=1+2\cdot 10^{n+1}+C_210^{n+3}$$
Basically, when $n\geq 2$, then $10^{n+3}\mid \binom{10}{i}2^{i}10^{ni}$ for $i\geq 2$. This is because when $i>2$, then $ni\geq n+3$ and when $i=2$, $ni\geq n+2$ and $10\mid\binom{10}{2}\cdot 2^2$.
