Higher Derivatives of Exponential Function

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

Let $$c$$ be a constant.

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

Proof
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 $$z = c x$$.

In this case, from Derivative of Exponential Function:
 * $$D{z} \left({\exp \left({z}\right)}\right) = \exp z$$

and so we don't even need induction to show that:
 * $$D^n{z} \left({\exp \left({z}\right)}\right) = \exp z$$

Hence the result.