16. Let the function $y=f(x)$ have the domain $\mathbf{R}$, and for $x>1$, and for any real numbers $x, y \in \mathbf{R}$, $f(x+y)=f(x) f(y)$ holds. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=f(0)$, and $f\left(a_{n+1}\right)=\frac{1}{f\left(-2-a_{n}\right)}(n \in \mathbf{N})$.
(1) Find the value of $a_{2003}$;
(2) If the inequality $\left(1+\frac{1}{a_{1}}\right)\left(1+\frac{1}{a_{2}}\right) \cdots\left(1+\frac{1}{a_{n}}\right) \geqslant k \cdot \sqrt{2 n+1}$ holds for all $n \in \mathbf{N}$, find the maximum value of $k$.
16. (1) Let $x=-1, y=0$, we get $f(-1)=f(-1) \cdot f(0), f(0)=1$. Therefore, $a_{1}=f(0)=1$. When $x>0$, $x_2 > x_1 > 0$, $f(x_2 - x_1) > 0$. Thus, $f(x_1) > f(x_2)$, and the function $y=f(x)$ is a monotonically decreasing function on $\mathbf{R}$. From $f(a_{n+1})=\frac{1}{f(-2-a_{n})}$, we get $f(a_{n+1}) \cdot f(-2-a_{n})=1$. Hence, $f(a_{n+1}-a_{n}-2)=f(0)$, $a_{n+1}-a_{n}-2=0$, $a_{n+1}-a_{n}=2(n \in \mathbf{N})$.
Therefore, $\{a_{n}\}$ is an arithmetic sequence with the first term 1 and common difference 2,
Thus, $a_{n}=2n-1$, $a_{2003}=4005$.
(2) From $\left(1+\frac{1}{a_{1}}\right)\left(1+\frac{1}{a_{2}}\right) \cdots\left(1+\frac{1}{a_{n}}\right) \geqslant k \sqrt{2 n+1}$ always holds,
we know $k \leqslant \frac{\left(1+\frac{1}{a_{1}}\right)+\left(1+\frac{1}{a_{2}}\right) \cdots\left(1+\frac{1}{a_{n}}\right)}{\sqrt{2 n+1}}$ always holds.
Let $F(n)=\frac{\left(1+\frac{1}{a_{1}}\right)\left(1+\frac{1}{a_{2}}\right) \cdots\left(1+\frac{1}{a_{n}}\right)}{\sqrt{2 n+1}}$,
we know $F(n)>0$, and $F(n+1)=\frac{\left(1+\frac{1}{a_{1}}\right)\left(1+\frac{1}{a_{2}}\right) \cdots\left(1+\frac{1}{a_{n}}\right)\left(1+\frac{1}{a_{n+1}}\right)}{\sqrt{2 n+3}}$,
also $\frac{F(n+1)}{F(n)}=\frac{2(n+1)}{\sqrt{4(n+1)^{2}-1}}>1$, i.e., $F(n+1)>F(n)$,
thus $F(n)$ is a monotonically increasing function with respect to $n$, $F(n) \geqslant F(1)=\frac{2}{3} \sqrt{3}$.
Therefore, $k \leqslant \frac{2}{3} \sqrt{3}$, i.e., the maximum value of $k$ is $\frac{2}{3} \sqrt{3}$.