Let $a$ be an integer number greater than one, ${k_i}_{1≤i≤r}$ and ${m_j}_{1≤j≤s}$ two strictly increasing sequences of integer numbers. Suppose that $$\sum_{i=1}^{r}\frac{1}{a^{k_i}}=\sum_{j=1}^{s}\frac{1}{a^{m_j}}$$
Prove that $r = s$ and for each $1 ≤ i ≤ r$, exists $1 ≤ j ≤ r$ such that $k_i = m_j$.
How to extend the result?
Let Suppose that $k_r\geq m_s$ and multiplying the equality by $a^{k_r}$ $$a^{k_r}\sum_{i=1}^{r}\frac{1}{a^{k_i}}=a^{k_r}\sum_{j=1}^{s}\frac{1}{a^{m_j}}$$
$$\sum_{i=1}^{r}{a^{k_r-k_i}}=\sum_{j=1}^{s}{a^{k_r-m_j}}$$
With $a=10$ and $\mod 9$ then $$\sum_{i=1}^{r}{1^{k_r-k_i}}=\sum_{j=1}^{s}{1^{k_r-m_j}}$$ $$\sum_{i=1}^{r}{1}=\sum_{j=1}^{s}{1}$$
Then $r=s$.
How to extend the result for any base $a$?
Maybe with $\mod a^{k_i}$ where $i=r,r-1,\cdots, 1$
Multiply by $a^{k_r-1}$. Then the left side turns into a non-integer, hence so must the right side, hence $m_s\ge k_r$. By symmetry, also $k_r\ge m_s$, hence $k_r=m_s$. This allows us to drop the last summand on both sides and proceed by induction.