If $a > 1$, $x >0$, show that $x^a - 1 \ge a(x-1)$. The question was taken from Serge Lang's "A First Course in Calculus". The particular section where this question was found is Exponents and Logarithms. 
If $a > 1$, $x >0$, show that $x^a - 1 \ge a(x-1)$. 
I let $f(x) = x^a - 1- ax + a$. Note that $f(0) = a - 1 > 0$ and if we can prove that $f' \ge 0$ for $x>0$ then the inequality holds. Now consider the first derivative $f'(x) = ax^{a-1} - a = a(x^{a-1} -1)$. It suffices to show that $x^{a-1} - 1 \ge 0$. 
I was tempted to say that since $a-1 > 0$, then $x^{a-1} > x^0 = 1$ for $x >0$ and therefore $x^{a-1} - 1 \ge 0$. However, I do not really know how true my previous assertion is.

From your $f'$ formula, it's easy to see $f$ is decreasing on $(0,1)$ and increasing on $(1,\infty)$, that means $f$ achieves minimum at $x=1$, and $f(1)=0$. Therefore $f(x)\geq 0$. 
