Iranian Mathematical Olympiad

Without loss of generality, suppose $s_1\ge r_1+1$. Let $d=\gcd (a,k)$, $a=d{a}_1$ and $k=d{k}_1$. By using this equation and rewriting the problem's equality it is obtained that $$\begin{array}{rl}\prod _{i=1}^n(d^{r_i}{a}_1^{r_i}+d{k}_1)& =\prod _{i=1}^n(d^{s_i}{a}_1^{s_i}+d{k}_1)\\ ⟹\prod _{i=1}^n(d^{r_i-1}{a}_1^{r_i}+k_1)& =\prod _{i=1}^n(d^{s_i-1}{a}_1^{s_i}+k_1)\end{array}$$ 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 $$\begin{array}{rl}& (d^{r_1-1}{a}_1^{r_1}+k_1)k_1^{n-1}\equiv k_1^n(\mathrm{mod}{d}^{r_1}{a}_1^{r_1+1})\\ ⟹\begin{array}{rl}& d^{r_1}{a}_1^{r_1+1}(\mathrm{mod}{d}^{r_1-1}{a}_1^{r_1}{k}_1^{n-1})\\ & d{a}_1(\mathrm{mod}{k}_1^{n-1})\end{array}& \end{array}$$ 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 $r_i=s_i$. ■