How can I prove that for any positive real number $C$, strictly greater than $1$, the function $$f(x)=C^x$$ is strictly increasing?

The definition of a strictly increasing function says that if $a<b$ then $f(a)<f(b)$.

Okay, suppose $a<b$. How to show that $C^a < C^b$?

My idea was to prove it by contradiction.

Let's say $C^a \ge C^b$. Taking $\log_c$ both sides (the fact the direction of inequality doesn't change follows from the fact that log is strictly increasing, so this proof doesn't change much actually):

$a\ge b$, a contradiciton, because we assumed $a<b$.

I'm actually wondering if it's possible to prove it without using derivatives.

1$ if $x>0$.– Tom-Tom Dec 16 '15 at 10:50