Power Function on Base between Zero and One is Strictly Decreasing/Positive Integer

Theorem
Let $a \in \R$.

Let $0 < a < 1$.

Let $f : \N \to \R$ be the real-valued function defined as:
 * $f \left({ n }\right) = a^n$

where $a^n$ denotes $a$ to the power of $n$.

Then $f$ is strictly decreasing.

Proof
For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle 0 < a < 1 \implies 0 < a^{n+1} < a^n$

Basis for the Induction
$P(1)$ is true, since $0 < a < 1 \implies 0 < a^2 < a^1$ by Real Number between Zero and One is Greater than Square.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:


 * $\displaystyle 0 < a < 1 \implies 0 < a^{k+1} < a^{k}$

Then we need to show:


 * $\displaystyle 0 < a < 1 \implies 0 < a^{k+2} < a^{k+1}$

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $0 < a < 1 \implies 0 < a^{n+1} < a^n$

Hence the result.