Given $n \in N$, prove that $f: [0, \infty) \rightarrow [0, \infty) f(x) = x^n$ is increasing without calculus (derivates) I have to prove this using $x_1 < x_2 \rightarrow f(x_1) < f(x_2)$.
I have no idea on how to do this, but what I have tried so far is:
Consider $x_1, x_2  \in [0, \infty)$ such that $x_1 < x_2 \rightarrow f(x_1) < f(x_2)$
$\rightarrow x_1^n < x_2^n$
I do not know what to do next
 A: You have
$$x_2^n - x_1^n = (x_2-x_1) (x_2^{n-1} + x_2^{n-2} x_1 + \cdots + x_2 x_1^{n-2}+ x_1^{n-1}) \gt 0$$ for $x_2 \gt x_1$.
A: We can use the result $a\le b\implies ax\le bx\quad\forall x\ge 0$
Note that since $0\le x_1<x_2$ by hypothesis then $x_2\neq 0$,
So we can divide and get $\  0\le a=\dfrac{x_1}{x_2}<1$ and multiplying by $a$ then $0\le a^2\le a<1$
And we can proceed by induction on $n$ to show $a^n\le a^{n-1}<1$, which we can retransform back into $x_1^n<x_2^n$.
A: Assumed that $n \in \Bbb{Z^+}.$
Alternative approach that is based on the binomial theorem:
$(a + b)^n = \sum_{k=0}^n \binom{n}{k}a^{n-k}b^k.$
$x_2 > x_1 \geq 0 \implies \exists b \in \Bbb{R^+}$ such that
$x_2 = (x_1 + b)$.
Therefore,
$$(x_2)^n - (x_1)^n = (x_1 + b)^n - (x_1)^n = \sum_{k=1}^n \binom{n}{k}(x_1)^{n-k}b^k > 0.$$
Note that the above summation skips the term corresponding to $k=0$.  Further, regardless of whether $x_1 = 0$ or $x_1 > 0$, all of the terms are non-negative, and the very last term (which is $b^n$) must be positive.
A: Say $x,y \in [0,\infty) $ and $x\lt y$. Then $$f(x) \lt f(y) \\ \iff x^n \lt y^n \\ $$
Since $\ln x$ is an increasing function, $$\iff n\ln x \lt n \ln y \\ \iff \ln x \lt \ln y \\ \iff x\lt y $$
But that’s of course true.
