Without loss of generality, suppose $s_1 \ge r_1 + 1$. Let $d = \gcd(a, k)$, $a = da_1$ and $k = dk_1$. By using this equation and rewriting the problem's equality it is obtained that
$$\begin{align*} \prod_{i=1}^{n} (d^{r_i} a_1^{r_i} + dk_1) &= \prod_{i=1}^{n} (d^{s_i} a_1^{s_i} + dk_1) \\ \implies \prod_{i=1}^{n} (d^{r_i-1} a_1^{r_i} + k_1) &= \prod_{i=1}^{n} (d^{s_i-1} a_1^{s_i} + k_1) \end{align*}$$
Now consider the latest two equations modulo $d^{r_1} a_1^{r_1+1}$. Since the sequences are strongly ascending, it is deduced that
$$\begin{align*} & (d^{r_1-1} a_1^{r_1} + k_1) k_1^{n-1} \equiv k_1^n \pmod{d^{r_1} a_1^{r_1+1}} \\ \implies \begin{aligned} & d^{r_1} a_1^{r_1+1} \pmod{d^{r_1-1} a_1^{r_1} k_1^{n-1}} \\ & da_1 \pmod{k_1^{n-1}} \end{aligned} \end{align*}$$
So $a_1 \mid k_1^{n-1}$, but $\gcd(a_1, k_1) = 1$. Therefore $a_1 = 1$ that results in $a \mid k$ and it's in contradiction with $a > k$. Therefore $s_1 = r_1$. After removing $s_1$, $r_1$ from the assumption and continuing the same thing for $r_2$, $s_2$ we take $r_2 = s_2$. So for each $i$ there is $r_i = s_i$. ■