8. Let $a>1$. If the equation $a^{x}=x$ has no real roots, then the range of the real number $a$ is $\qquad$ .
8. $a>\mathrm{e}^{\frac{1}{e}}$.

From the graphs of the functions $y=a^{x}$ and $y=x$, we know that if $a>1$ and $a^{x}=x$ has no real roots, then $a^{x}>x$ always holds.
Let $f(x)=a^{x}-x$. Then
$f^{\prime}(x)=a^{x}(\ln a)-1>0 \Rightarrow x>-\log _{a}(\ln a)$.
Therefore, $f(x)=a^{x}-x$ is decreasing on the interval $\left(-\infty,-\log _{a}(\ln a)\right)$ and increasing on the interval $\left(-\log _{a}(\ln a),+\infty\right)$. Thus, $f(x)$ achieves its minimum value at $x=-\log _{a}(\ln a)$, i.e.,
$$
\begin{array}{l}
f(x)_{\min }=f\left(-\log _{a}(\ln a)\right) \\
=a^{-\log _{a}(\ln a)}-\left(-\log _{a}(\ln a)\right)>0 \\
\Rightarrow \frac{1}{\ln a}-\left(-\log _{a}(\ln a)\right)>0 . \\
\text { Also, } \frac{1}{\ln a}=\log _{a} \mathrm{e},-\log _{a}(\ln a)=\log _{a} \frac{1}{\ln a} \\
\Rightarrow \log _{a} \mathrm{e}>\log _{a} \frac{1}{\ln a} \Rightarrow \ln a>\frac{1}{\mathrm{e}} \Rightarrow a>\mathrm{e}^{\frac{1}{a}} .
\end{array}
$$