Proof that $3^{10^n}\equiv 1\pmod{10^n},\, n\ge 2$ This should be rather straightforward, but the goal is to prove that
$$3^{10^n}\equiv 1\pmod{10^n},\, n\ge 2.$$
A possibility is to use
$$\begin{align*}3^{10^{n+1}}-1&=\left(3^{10^n}-1\right)\left(1+\sum_{k=1}^9 3^{10^n k}\right)\\&=\left(3^{10^n}-1\right)\left(3^{9\cdot 10^n}-1+3^{8\cdot 10^n}-1+\cdots +3^{10^n}-1+10\right),\end{align*}$$
but I don't see how this proves the equality in the question. I got only
$$3^{10^{n}}\equiv 1\pmod{3^{10^{n-1}}-1}.$$
Perhaps someone here can explain.

Use induction
Basis
$$3^{100}\equiv (3^{10})^{10}$$
$$\equiv 59049^{10}$$
$$\equiv 49^{10}$$
$$\equiv 2401^5$$
$$\equiv 1\pmod {100}$$
Induction hypothesis
$$\frac{3^{10^{n}}-1}{10^n}\in\mathbb Z$$
Inductive step
$$\frac{3^{10^{n+1}}-1}{10^{n+1}}$$
$$=\frac{3^{10^{n}}-1}{10^n}\times \frac{1+\sum_{k=1}^9 3^{10^nk}}{10}$$
But of the terms multiplied are integers, as the first one directly follows from induction hypothesis and second one by $$\forall 1\leq k\leq 9, 10|10^n|(3^{10^n}-1)|(3^{10^nk}-1)$$
$$\Longrightarrow 3^{10^nk}\equiv 1\pmod {10}\forall 1\leq k\leq 9$$
$$\Longrightarrow 1+\sum_{k=1}^9 3^{10^nk}\equiv 0\pmod {10}$$
Hence proved
