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.

The identity you used leads to a proof of the desired result by induction as follows. So suppose that $10^n$ divides $3^{10^n}-1$ for all $n=2,\ldots,N$. Applying the identity yields
$$
\frac{3^{10^{N+1}}-1}{3^{10^N}-1}=10 + \sum_{k=1}^9(3^{k\cdot 10^N}-1),
$$
and the right side is a multiple of $10$ since each term in the sum is a multiple of $3^{10^N}-1$.
Thus, since $10^N$ divides $3^{10^N}-1$ by the inductive hypothesis, we see that $10^{N+1}$ divides $3^{10^{N+1}}-1$ and this completes the inductive step. All that is left is to check the base case $n=2$, that $3^{100}-1$ is divisible by $100$. This is straightforward using standard modular arithmetic techniques, for instance by checking modulo $4$ and modulo $25$ separately.
