1. When $n \in \mathbf{N}$, and $n>2$, the graphs of the functions $y=\log _{n-1} x$ and $y=\log _{n}(x+1)$ intersect at all points where
2. $(n-1,1)$.
$$
\left\{\begin{array} { l } 
{ y = \operatorname { log } _ { n - 1 } x , } \\
{ y = \operatorname { log } _ { n } ( x + 1 ) }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=(n-1)^{y}, \\
x+1=n^{y} .
\end{array}\right.\right.
$$

Eliminating $x$ yields
$$
\begin{array}{l}
1+(n-1)^{y}=n^{y} \\
\Leftrightarrow f(y)=\left(\frac{1}{n}\right)^{y}+\left(\frac{n-1}{n}\right)^{y}=1 .
\end{array}
$$

Since $0<\frac{1}{n} 、 \frac{n-1}{n}<1$, $f(y)$ is a decreasing function.
Also, $f(1)=\frac{1}{n}+\frac{n-1}{n}=1$, thus, $y=1$ is the unique solution to $f(y)=1$. At this point, $x=n-1$.
Therefore, the graphs of the two functions have a unique intersection point $(n-1,1)$.