<s>[INST] Show that $(x-1)^2$ is a factor of $x^n -nx +n-1$ 
Show that $(x-1)^2$ is a factor of $x^n -nx +n-1$

By factor theorem we know that $(x-a)$ is a factor of $f(x)$ if $f(a)=0$. 
In this case, $f(x)=x^n -nx +n-1 \implies f(1)=0$
Hence we conclude that $(x-1)$ is a factor. From hereon, how can I say that $(x-1)^2$ is a factor?  
Can we approach the problem without calculus approach? This problem was taken from a book of pre-calculus algebra.
 [/INST] Here is another elementary way using the binomial theorem


*

*$(1+y)^n = \sum_{k=0}^n\binom{n}{k}y^k$
Set $\boxed{y=x-1}$ and note that


*

*$(x-1)^2$ is a factor of $p(x) = x^n -nx +n-1$ if and only if $y^2$ is a factor of $p(y+1)$
Hence,
\begin{eqnarray*} p(y+1)
& = & (1+y)^n - (1+y)n + n-1\\
& = & 1+ny +\sum_{k=2}^n\binom{n}{k}y^k -n-ny+n-1 \\
& = & y^2\sum_{k=2}^n\binom{n}{k}y^{k-2}
\end{eqnarray*}
Done.
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