Higher Derivatives of Exponential Function

Theorem
Let $\exp x$ be the exponential function.

Then:
 * $\map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$

Proof
Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
 * $\map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$

Basis for the Induction
$\map P 1$ is true, as this is the case proved in Derivative of Exponential Function:
 * $\map {\dfrac \d {\d x} } {\exp x} = \exp x$

This is our basis for the induction.

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

So this is our induction hypothesis:
 * $\map {\dfrac {\d^k} {\d x^k} } {\exp x} = \exp x$

Then we need to show:
 * $\map {\dfrac {\d^{k + 1} } {\d x^{k + 1} } } {\exp x} = \exp x$

Induction Step
This is our induction step:

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

Therefore:
 * $\forall n \in \N_{>0}: \map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$