Let I $=\int 7^{7^{7^{x}}} 7^{7^{x}} 7^{x} \mathrm{dx}$
Let $y=7^{7^{7^{x}}}$
$\log y=7^{7^{x}}(\log 7)$
Differentiating w.r.t. x, we get
$$
\frac{1}{y} \frac{d y}{d x}=(\log 7) \frac{d}{d x}\left(7^{7^{x}}\right)
$$
Let $z=7^{7^{x}} \Rightarrow \log z=7^{x} \log 7$
Differentiating w.r.t. $\mathrm{x}$, we get
$$
\begin{aligned}
& \frac{1}{z} \frac{d z}{d x}=7^{x}(\log 7)^{2} \\
\therefore \quad & \frac{d}{d x}\left(7^{7^{x}}\right)=7^{7^{x}} 7^{x}(\log 7)^{2} \Rightarrow \frac{d y}{d x}=7^{7^{7}} 7^{7^{x}} 7^{x}(\log 7)^{3} \\
\therefore \quad I=& \int \frac{d}{d x}\left(7^{7^{7^{x}}}\right) \times \frac{1}{(\log 7)^{3}} d x=\frac{7^{7^{x}}}{(\log 7)^{3}}+C
\end{aligned}
$$