1. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=\mathrm{e}, a_{2}=\mathrm{e}^{3}, \\
\mathrm{e}^{1-k} a_{n}^{k+2}=a_{n+1} a_{n-1}^{2 k}\left(n \geqslant 2, n \in \mathbf{Z}_{+}, k \in \mathbf{R}_{+}\right) . \\
\text {Find } \prod_{i=1}^{2017} a_{i} . \quad \text { (Hao Hongbin) }
\end{array}
$$
(Hao Hongbin)
1. For $\mathrm{e}^{1-k} a_{n}^{k+2}=a_{n+1} a_{n-1}^{2 k}$, taking the natural logarithm on both sides gives
$$
\begin{array}{l}
1-k+(k+2) \ln a_{n}=\ln a_{n+1}+2 k \ln a_{n-1} \\
\Rightarrow 1+\ln a_{n+1} \\
\quad=(k+2)\left(1+\ln a_{n}\right)-2 k\left(1+\ln a_{n-1}\right) .
\end{array}
$$

Let $b_{n}=1+\ln a_{n}$.
$$
\begin{array}{l}
\text { Then } b_{n+1}=(k+2) b_{n}-2 k b_{n-1}(n \geqslant 2) \\
\Rightarrow b_{n+1}-2 b_{n}=k\left(b_{n}-2 b_{n-1}\right)(n \geqslant 2) . \\
\text { Also, } b_{1}=1+\ln a_{1}=1+\ln \mathrm{e}=2, \\
b_{2}=1+\ln a_{2}=1+\ln \mathrm{e}^{3}=4,
\end{array}
$$

Then $b_{n+1}-2 b_{n}=0 \Rightarrow b_{n}=2^{n}\left(n \in \mathbf{Z}_{+}\right)$
$$
\Rightarrow b_{n}=1+\ln a_{n}=2^{n} \Rightarrow a_{n}=\mathrm{e}^{2 n-1} \text {. }
$$

Let $s=\sum_{i=1}^{2017}\left(2^{i}-1\right)$.
Thus $\prod_{i=1}^{2017} a_{i}=\mathrm{e}^{s}=\mathrm{e}^{2^{2018}-2019}$.