Higher Derivatives of Exponential Function

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

Let $c$ be a constant.

Then:
 * $D^n_x \left({\exp x}\right) = \exp x$

Corollary

 * $D^n_x \left({\exp \left({c x}\right)}\right) = c^n \exp \left({c x}\right)$

Proof
Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $D^n_x \left({\exp x}\right) = \exp x$

Basis for the Induction
$P(1)$ is true, as this is the case proved in Derivative of Exponential Function:
 * $D_x \left({\exp x}\right) = \exp x$

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:
 * $D^k_x \left({\exp x}\right) = \exp x$

Then we need to show:
 * $D^{k+1}_x \left({\exp x}\right) = \exp x$

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:
 * $\forall n \in \N^*: D^n_x \left({\exp x}\right) = \exp x$

Proof of Corollary
This follows directly from Derivatives of Function of ax + b:
 * $D^n_x \left({f \left({a x + b}\right)}\right) = a^n D^n_{z} \left({f \left({z}\right)}\right)$

where $z = a x + b$.

Here we set $a = c$ and $b = 0$ so that:
 * $D^n_x \left({f \left({c x}\right)}\right) = c^n D^n_{z} \left({f \left({z}\right)}\right)$

where $z = c x$.

Then from the main result:
 * $D^n_z \left({\exp \left({z}\right)}\right) = \exp z$

Hence the result.