Proving $x^\alpha-\alpha x \le 1- \alpha $

Let $f(x)=x^\alpha-\alpha x-\alpha+1$ for $x\geq 0$, where $\alpha<1$, so $f'(x)=\alpha x^{\alpha-1}-\alpha$ and $f''(x)=\alpha(\alpha-1)x^{\alpha-2}$. Then $f'(x)=0\Rightarrow x=1$ with $f''(1)=\alpha(\alpha-1)\leq 0$. This implies that $f$ has a local maximum at $x=1$ and so $f(1)=2(1-\alpha)\geq1-\alpha=f(0)$ . Also we have $\displaystyle \lim_{x\to\infty}f(x)=-\infty$. Since $f$ is continuous for $x\geq 0$, considering above values we see that $f$ has a maximum at $x=1$ and equals $f(1)=2(1-\alpha)$. Therefore we have $f(x)\leq f(1)$ for all $x\geq 0$ and this leads to the desired inequality.

The problem is too easy once one rewrites it as $x^{\alpha} - 1 \leq \alpha(x - 1)$. If $f(x) = x^{\alpha}$, then $f'(x) = \alpha x^{\alpha - 1}$ and then by Mean Value Theorem we have $$x^{\alpha} - 1 = f(x) - f(1) = (x - 1)f'(c) = \alpha(x - 1)c^{\alpha - 1}$$ where $c$ is between $1$ and $x$. Clearly we need to consider cases when $x < 1$ and $x > 1$. For $x = 1$ the result is trivial. Supposing that $x > 1$ we need to ensure that $c^{\alpha - 1} < 1$ which is practically obvious as $1 < c < x$ and $\alpha - 1 < 0$. Same way if $x < 1$ then we need to establish that $c^{\alpha - 1} > 1$ and again this is obvious because then $x < c < 1$.