Prove $x^n -1 \ge (x-1)^n \ \forall n \in \mathbb{N}\ \land \ \forall x\ \ge 1 $ Does my proof make sense?

Problem: prove $x^n -1 \ge (x-1)^n \ \forall n \in \mathbb{N}\  \land \ \forall x\ \ge 1 $
Base step (P(1)):$\  \ \ \ \ \ \ \ x-1 \ge x-1$
Hypotesis (P(n)): $ \ \ \ \ \ x^n -1 \ge (x-1)^n$
Thesis (P(n+1)):$ \ \ \ \ \ \ \ \ x^{n+1}-1 \ge (x-1)^{n+1}$
Induction step:
$$x^{n+1}-1 = x^n \cdot x -1 = x \cdot x^n -1 \ge x \cdot (x-1)^n \ge (x-1)^{n+1}$$
where: $x \cdot (\not x\not-\not1\not)^n \ge \not(\not x\not-\not1\not)^n \cdot  (x-1)$ so $x \ge x-1$

 A: Well if you have $x^n-1\ge (x-1)^n$ then multiply it by $(x-1)$ to get:
$$x^{n+1}+1-x-x^n=(x^n-1)(x-1)\ge (x-1)^{n+1}$$
and the LHS is less than $x^{n+1}-1$.
A: $$x^{n+1}-1$$
$$=(x-1+1)x^n-1$$
$$=(x-1)x^n + x^n-1$$
$$\ge (x-1)x^n + (x-1)^n$$
$$\ge (x-1)x^n$$
$$\ge (x-1)^{n+1}$$
Your solution isn't valid as $x\cdot x^n-1\ne x(x^n-1)\;\;\forall x\in\mathbb{R}$.
A: $(x-1)^n=\binom{n}{0}(-1)^n+\binom{n}{1}(-1)^{n-1}x+\cdots+\binom{n}{n}x^n$ by the Binomial Theorem.
Evaluating the above at $x=1$ we see
$\sum\limits_{i=0}^n(-1)^{n-i}\binom{n}{i}=0, \sum\limits_{i=0}^{n-1}(-1)^{n-i}\binom{n}{i}+\binom{n}{n}=0 \therefore (\color{red}{*})\sum\limits_{i=0}^{n-1}(-1)^{n-i}\binom{n}{i}=-1 $ because $\binom{n}{n}=1$.
Let
$f(x)=(x-1)^n-x^n=\binom{n}{0}(-1)^n+\binom{n}{1}(-1)^{n-1}x+\cdots +\binom{n}{n-1}(-1)^1x^{n-1}$,
$\frac{d}{dx}[f(x)]=\frac{d}{dx}[(x-1)^n-x^n]=n(x-1)^{n-1}-nx^{n-1}=n((x-1)^{n-1}-x^{n-1})\leq 0$ for $x\geq1$.
Note that $f(1)=-1$ by above $(\color{red}{*})$, and $f$ is decreasing, thus $f(x)\leq -1$ for $x\geq 1$.
$\therefore (x-1)^n=f(x)+x^n\leq -1+x^n$ for $x\geq 1, n\in \mathbb{N}$.
A: Here's a proof without induction that allows $n$ to be any real number $\gt 1.$ Let $f(x)= x^n -1$ and $g(x)= (x-1)^n.$
Observe that $f'(x)= nx^{n-1}$ and $g'(x)= n(x-1)^{n-1}.$ Since $f(1)=g(1)$ and $f'(x) \gt g'(x) \text{ whenever } x \gt 1,$ it follows that $f(x) \gt g(x) \text{ whenever } x \gt 1.$
