Let's find the highest degree expression of $x$ such that

$$
n x^{n+1}-(n+1) x^{n}+1
$$

and $x^{n}-n x+n-1$

are divisible.
Add up the two polynomials, the sum is

$$
n(x-1)\left(x^{n}-1\right)
$$

Subtract the $n(x-1)$ multiple of (2) from (3), then we get $n^{2}(x-1)^{2}$. Therefore, the common factor can only be $(x-1)^{2}$. However, this is indeed a common factor, because (1) can be written as

$$
\begin{aligned}
& n x^{n+1}-(n+1) x^{n}+1=(x-1)\left[n x^{n}-\left(1+x+x^{2}+\ldots+x^{n-1}\right)\right]= \\
& \quad=(x-1)\left[\left(x^{n}-1\right)+\left(x^{n}-x\right)+\left(x^{n}-x^{2}\right)+\ldots+\left(x^{n}-x^{n-1}\right)\right]
\end{aligned}
$$

(2) is

$$
\begin{gathered}
x^{n}-n x+n-1=(x-1)\left[1+x+x^{2}+\ldots+x^{n-1}-n\right]= \\
\quad=(x-1)\left[(x-1)+\left(x^{2}-1\right)+\ldots+\left(x^{n-1}-1\right)\right]
\end{gathered}
$$

However, it is clear that $(x-1)$ can be factored out from both expressions in the square brackets if $n \geq 2$. If, however, $n=1$, then (2) is identically 0, or $0 \cdot(x-1)^{2}$, and (1) is precisely $(x-1)^{2}$.