Determining if a function is strictly increasing for all natural numbers on a specific interval I am having some difficulties with tackling the following problem:
Show that $f(x)=x^n$ (where n ∈ $\mathbb{N}$) is strictly increasing on the interval $[0, ∞)$.
I am familiar with the following definition: Strictly increasing means if $x_1 &lt; x_2$ then $f(x_1) &lt; f(x_2)$. So, I started with this definition. However, I am not sure if this is supposed to be proved using the mathematical induction method. I tried to split it into two cases: an even and an odd exponent, but this didn't really lead me anywhere, unfortunately.
If you know how to use induction, then proceed like this: $f(x)=x^1=x$ is clearly strictly increasing, because if $0 \leq x&lt;y$, then $f(x)=x&lt;y=f(y)$. Now assume the result is true for $n=1,2,\ldots,k$. We need show this is true for $n=k+1$. 

Suppose $0 \leq x&lt;y$. Let $f(x)=x^{k+1}$ and $g(x)=x^k$. By assumption, we know that $g(x)$ is increasing so that $g(x)= x^k&lt;y^k=g(y)$. But
$$
f(x)= x^{k+1}= x \cdot x^k= x \cdot g(x) \leq y \cdot g(x) &lt; y \cdot g(y)= y \cdot y^k= y^{k+1}= f(y)
$$
But then $f(x)$ is increasing.
I just taught this today.

Let $0\leq x_1&lt;x_2$.

Notice that

$x_2^n-x_1^n=(x_2-x_1)(\sum_{i=0}^{n-1}x_1^i x_2^{n-1-i})$ is positive since both factor are. Hence $x_2^n&gt;x_1^n$ and the function is increasing.