Multiplicity of root $1$ of $nX^{n+2}-(n+2)X^{n+1}+(n+2)X-n$? I'm solving this homework problem:

Show that $1$ is a root of the polynomial $nX^{n+2}-(n+2)X^{n+1}+(n+2)X-n$. Determine its multiplicity.

I believe the answer is $n+2,$ and currently working on proving this. My intuition stems from the fact that when $n=1, P(x)=(x-1)^3, so$ 1 is a polynomial of multiplicity $3.$
I attempted this: calling the polynomial $P(x),$ we see that $P(1)=0,$ hence $1$ is a root for sure. To determine its multiplicity, I'm thinking about:
Taking successive derivatives $P'(1), P''(1) \dots P^{(n+3)}(1) $ of the polynomial and showing that all but the last one $P^{(n+3)}(1)$ vanishes. But I'm a bit skeptical because the antiderivative of a polynomial doesn't always have multiplicity one more than the original polynomial, which forms the basis/idea of the method I'm thinking of. For example $Q(x):=2x$ has the root $0$ of multiplicity $1,$ but the antiderivative $x^2+1$ doesn't have any real root.
So my question is: in order to answer the question in the image, what is the exact theorem should we use? Is it something like this?
Proposed theorem: If $Q(x)$ has the real root $a$ of order $k,$ then its antiderivative with constant of integration $0$ has the real root $a$ of order $k+1.$
The above seems to be true, and if yes, can we use this to show that the successive derivatives $P'(1), P''(1) \dots P^{(n+2)}(1) $ of the polynomial vanish but $P^{(n+2)}(1) $ doesn't vanish, and this'll show that $1$ is a root of multiplicity $n+2.$ Is my idea correct?
ADDENDUM/EDIT: Can we arrive at the proof that $1$ is a root of multiplicity $n+2$ by using induction? At $n=1,$ the multiplicity is $1+2=3,$ so the induction can start, and now we just need to show the induction step. Will this work? P.S. As per Dietrich's answer, the multiplicity seems to be $3$ irrespective of $n.$ So can we just prove this using the derivatives above or by induction?

We may also look at what the coefficients tell us:  the polynomial $$ \ p(x) \ \ =  \ \ n·x^{n+2} \ - \ (n+2)·x^{n+1} \ + \ 0·x^n \ + \ \ldots \ + \ 0·x^2 \ + \ (n+2)·x \ - \ n \ \ $$  is anti-palindromic, which is to say that the coefficients "read left-to-right" are the negative of the sequence "read right-to-left".   Consequently, $ \ x \ = \ 1 \ $ must be a zero of such a polynomial, as the terms in $ \ p(1) \ $ then "cancel in pairs".  We also see from the Rule of Signs that $ \ p(x) \ $ has three "sign-changes", indicating that it has either three or one positive real zeroes, regardless of the parity of $ \ n \ \ . $  Since for $ \ n \ $ even,
$$ \ p(-x) \ \ =  \ \ n·x^{n+2} \ + \ (n+2)·x^{n+1} \ + \ 0·x^n \ + \ \ldots \ + \ 0·x^2 \ - \ (n+2)·x \ - \ n \ \ $$
has one "sign-change", there is also one negative real root; for $ \ n \ $ odd,
$$ \ p(-x) \ \ =  \ \ -n·x^{n+2} \ - \ (n+2)·x^{n+1} \ + \ 0·x^n \ + \ \ldots \ + \ 0·x^2 \ - \ (n+2)·x \ - \ n \ \ $$
has no "sign-changes", so  in this case, $ \ p(x) \ $ has  no negative real roots.
Upon employing  polynomial/synthetic division  for $ \ x \ = \ 1 \ \ , $ we obtain the "reduced" polynomial
$$ p_1(x) \ \ = \ \ n·x^{n+1} \ \underbrace{- \ 2·x^n \ - \ 2·x^{n-1} \ + \ \ldots \ - \ 2·x^2 \ - \ 2·x}_{n \ \text{terms}} \ + \ n \ \ , $$
which is palindromic, thus having the property that if $ \ r \ $ is a zero, then $ \ \frac{1}{r} \ $ is also.  We observe that $ \ x \ = \ 1 \ $ is a zero of this polynomial as well, so it also has a zero $ \ x \ = \ \frac11 \ = \ 1 \ \ , \  $  which establishes that the three positive real zeroes of $ \ p(x) \ $ are $ \ 1 \ $ with multiplicity $ \ 3 \ \ . $
With $ \ n \ $ odd then, there are these three real zeroes with the remaining $ \ (n - 1) \ $ [even] zeroes forming complex-conjugate pairs.  For $ \ n \ $ even, the anti-palindromic character of $ \ p(x) \ $ implies that $ \ x \ = \ -1 \ $ is also a zero, giving us the expected three positive and one negative real zeroes and $ \ (n - 2) \ $ [even] complex-conjugate zeroes.
It might be mentioned, by Viete's relations, that as the product of all $ \ (n + 2) \ $ zeroes of  $ \ p(x) \ $ is $ \ +1 \ $ for $ \ n \ $ odd and $ \ -1 \ $ for $ \ n \ $ even, the product of the complex zeroes alone is always $ \ +1 \ \ . $  This means that the complex conjugate zeroes are in pairs $ \ r \ , \ \overline{r} \ = \ \frac{1}{r} \ \ , $ so all of the complex zeroes have unit modulus (so in fact all of the zeroes do).
We can also demonstrate the triple multiplicity by polynomial division.  Dividing $ \ p_1(x) \ $ by $ \ (x - 1) \ $ produces
$$ p_2(x) \ \ = \ \ n·x^n \  + \ \underbrace{(n - 2)·x^{n-1} \ + \ (n - 4)·x^{n-2} \ + \ \ldots \ + \ (n - [2n - 2] )·x \ + \ (n - 2n)}_{n \ \text{terms}} $$ $$  = \ \ n·x^n \ + \ (n - 2)·x^{n-1} \ + \ (n - 4)·x^{n-2} \ + \ \ldots \ - \ (n - 4)·x^2 \ - \ (n -2)·x \ - \ n \ \ , $$
which is anti-palindromic, and so has $ \ x \ =  \ 1 \ $ as a zero.  One further division by $ \ (x - 1) \ $ yields
$$ p_3(x) \ \ = \ \ n·x^{n-1} \  + \ \underbrace{2·(n - 1)·x^{n-2} \ + \ 3·(n - 2)·x^{n-3} \ + \ \ldots \ + \ (n - 1)·(n - [n - 2] )·x \ + \ n·(n - [n - 1])}_{(n - 1) \ \text{terms}} $$ $$  = \ \ n·x^{n - 1} \ + \ 2·(n - 1)·x^{n-2} \ + \ 3·(n - 2)·x^{n-3} \ + \ \ldots $$ $$ + \ 3·(n - 2)·x^2 \ + \ 2·(n - 1)·x \ + \ n \ \ . $$
Since the exponents in the terms are non-negative integers, $ \ p_3(x) \ $ has only positive coefficients, so $ \ x \ = \ 1 \ $ cannot be one of its zeroes.  Hence, $ \ p(x) \ $ has a factor limited to $ \ (x - 1)^3 \ \ . $
