We have, $f_{1}(x)=e^{x}$
$$
\begin{array}{r}
f_{2}(x)=e^{f_{1}(x)} \\
\cdots \cdot \cdots \cdots \\
\cdots f_{n+1}(x)=e^{f_{n}(x)}
\end{array}
$$
$f_{n}(x)=e^{f_{n-1}(x)}$
On taking log both sides, we get
$\quad \log \left\{f_{n}(x)\right\}=f_{n-1}(x) \log e$
$\Rightarrow \quad \frac{d}{d x} \log \left(f_{n}(x)\right)=\frac{d}{d x} f_{n-1}(x) \quad(\because \log e=1)$
$\Rightarrow \quad \frac{1}{f_{n}(x)} \frac{d}{d x} f_{n}(x)=f_{n-1}^{\prime}(x)$
$\Rightarrow \quad \frac{d}{d x} f_{n}(x)=f_{n}(x) f_{n-1}^{\prime}(x) \quad \ldots$ (i)
Now, $\quad f_{1}^{\prime}(x)=e^{x}$
$\begin{array}{l}\text { and } \quad f_{2}(x)=e^{f_{1}(x)} \\ \Rightarrow \log f_{2}(x)=f_{1}(x) \log e=f_{1}(x) \\ \Rightarrow \quad \frac{1}{f_{2}(x)} \cdot f_{2}^{\prime}(x)=f_{1}^{\prime}(x) \\ \Rightarrow \quad \begin{aligned} f_{2}^{\prime}(x) &=f_{2}(x) \cdot f_{1}^{\prime}(x) \\=f_{2}(x) \cdot e^{x} &\left(\because f_{1}^{\prime}(x)=e^{x}\right) \end{aligned} \\ =f_{2}(x) \cdot f_{1}(x) & {\left[\because e^{x}=f_{1}(x)\right] \ldots(\mathrm{ii})}\end{array}$
From Eq. (i). $\frac{d}{d x} f_{n}(x)=f_{n}(x) \cdot f_{n-1}(x) \ldots f_{1}(x)[\mathrm{using} \mathrm{Eq}$