Example 11 Try to find the remainder when $10^{10}+10^{10^{2}}+10^{10^{3}}+\cdots+10^{10^{10}}$ is divided by 7.
Let $A=10^{10}+10^{10^{2}}+10^{10^{3}}+\cdots+10^{10^{10}}$.
First, we prove that if $6 \mid n-r$, then $7 \mid 10^{n}-10^{r},(n>r)$.
In fact,
$$
10^{n}-10^{r}=10^{r}\left(10^{n-r}-1\right)=10^{r}\left(10^{6 k}-1\right)=\left(\left(10^{6}\right)^{k}-1\right) \cdot 10^{r} \text {. }
$$

Since $10^{6} \equiv 1(\bmod 7)$, we have
$$
\left(10^{6}\right)^{k}-1 \equiv 0(\bmod 7) \text {, }
$$

which means $7 \mid 10^{n}-10^{r}$.
On the other hand, if $6 \mid 10^{k}-10(k \geqslant 1)$, then
$A-10 \cdot 10^{10}+10 \cdot 10^{10}$
$=\left(10^{10}-10^{10}\right)+\left(10^{10^{2}}-10^{10}\right)+\cdots+\left(10^{10^{10}}-10^{10}\right)+10 \cdot 10^{10}$.
Since $6 \mid 10^{k}-10$, we have
$7 \mid 10^{10^{k}}-10^{10}$.
Therefore,
$$
\begin{aligned}
A & \equiv 10 \cdot 10^{10} \\
& =10^{11} \\
& =(7+3)^{11} \\
& \equiv 3^{11} \\
& \equiv 3^{5} \cdot 3^{6}(\bmod 7) .
\end{aligned}
$$

Since $3^{5} \equiv 5(\bmod 7)$,
$$
3^{6} \equiv 1(\bmod 7),
$$

we have $A \equiv 5(\bmod 7)$.
Thus, the remainder when $A$ is divided by 7 is 5.