$T^n - T + 1$ is separable over a field $k$ unless $\mathrm{char}(k)$ divides $n^n - (n - 1)^{n - 1}$ for all $n \geq 2$

Question

I am trying to prove the statement:

$T^n - T + 1$ is separable over a field $k$ unless $\mathrm{char}(k)$ divides $n^n - (n - 1)^{n - 1}$ for all $n \geq 2$

My first thought was to prove this using induction on $n$. For $n = 2$, $T^2 - T + 1 = \frac{(2T - 1)}{4} (2T - 1) + \frac{3}{4}$ and thus $T^2 - T + 1$ and its derivative are coprime unless $3 \equiv 0$, i.e. $\mathrm{char}(K) = 3$. (What about when $char(K) = 4$, may we conclude directly that $T^2 - T + 1$ and its derivative are coprime?)

Now, for a general $n \geq 2$, similar reasoning leads to the following:

$T^n - T +1 = (nT^{n-1} - 1)(\frac{T}{n} - \frac{n - 1}{n^2 T^{n - 2}}) + \frac{n^2 T^{n - 2} - n + 1}{n^2 T^{n - 2}}$.

Then $T^n - T +1$ and its derivative would be coprime unless $n - 1 \equiv n^2 T^{n-2}$, which doesn't seem related to what I have to prove.

I'd appreciate any remarks or different methods to prove the statement.

Answer 1

Hint: Let $f(T)=T^n-T+1$. Then $f'(T)=nT^{n-1}-1$ and $$nf(T)-Tf'(T)=n(T^n-T+1)-T(nT^{n-1}-1)=-(n-1)T+n$$

So what can be the GCD of $f(T)$ and $f'(T)$?