8. Given the function $f(x)=\sqrt{1+x \sqrt{1+(x+1) \sqrt{1+(x+2) \sqrt{1+(x+3) \sqrt{\cdots}}}}}(x>1$, $\left.x \in \mathbf{N}^{*}\right)$, then $f(2008)=$ $\qquad$
8. 2009 Detailed Explanation: From $f(x)=\sqrt{1+x \sqrt{1+(x+1) \sqrt{1+(x+2) \sqrt{1+(x+3) \sqrt{\cdots}}}}}\left(x>1, x \in \mathbf{N}^{*}\right)$ we get $f^{2}(x)=1+x f(x+1)$. Conjecture $f(x)=k x+b(k \neq 0)$, substituting into the equation yields $k=1, b=1$. By limit theory (omitted) we know that $f(x)$ has only this expression. From $f(x)=x+1\left(x>1, x \in \mathbf{N}^{*}\right)$ we get $f(2008)=2009$.