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.