minecraftfaq wrote: Define sequence $(a_n):a_1=\text{e},a_2=\text{e}^3,\text{e}^{1-k}a_n^{k+2}=a_{n+1}a_{n-1}^{2k}$ for all $n\geq2$, where $k$ is a positive real number. Find $\prod_{i=1}^{2017}a_i$. We get $a_n=e^{2^n-1}$ (easy direct computation writing $a_n=e^{b_n-1}$ or direct check) So $\prod_{i=1}^{2017}a_i=e^{\sum_{i=1}^{2017}(2^i-1)}=\boxed{e^{2^{2018}-2019}}$