7. Calculate $\sqrt[n]{\underbrace{99 \cdots 9}_{n \uparrow} \times \underbrace{99 \cdots 9}_{n \uparrow}+1 \underbrace{99 \cdots 9}_{n \uparrow}}(n \geqslant 2$, $n \in \mathbf{N})$ 的值为

The value of $\sqrt[n]{\underbrace{99 \cdots 9}_{n \uparrow} \times \underbrace{99 \cdots 9}_{n \uparrow}+1 \underbrace{99 \cdots 9}_{n \uparrow}}(n \geqslant 2$, $n \in \mathbf{N})$ is
7. 100 .
$$
\begin{array}{l}
\sqrt[n]{\underbrace{99 \cdots 9}_{n \uparrow} \times \underbrace{99 \cdots 9}_{n \uparrow}+1 \underbrace{99 \cdots 9}_{n \uparrow}} \\
=\sqrt[n]{\left(10^{n}-1\right)^{2}+2 \times 10^{n}-1} \\
=\sqrt[n]{10^{2 n}}=100 .
\end{array}
$$