Two Proofs Of The Unique Self-Differentiation Property of $Ce^x$

Ever since I learned elementary calculus, I have been wondering about the proof of the unique self-differentiation of $Ce^x$, $C$ a constant. One of the proofs below I was led to by a calculus textbook; another I figured out by myself.

$1$: Using the Mean Value Theorem

Let $f(x)$ be a function so that $f(x)=f'(x)$. Define a new function

$g(x)=\frac{f(x)}{e^x} (1)$

Now, find $g'(x)$:

$g'(x)=\frac{e^xf'(x)-e^xf(x)}{e^{2x}}$

$g'(x)=\frac{e^x(f'(x)-f(x))}{e^{2x}}$

But our base assumption is that $f(x)=f'(x)$, or that $f'(x)-f(x)=0$. In other words,

$g'(x)=0$

Now, the Mean Value Theorem implies that only constant functions have zero derivative. This means that $g(x)=C$ for some constant $C$. Plugging this into $(1)$ yields:

$C=\frac{f(x)}{e^x}$

$f(x)=Ce^x$

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$2$: Using the Taylor Expansion of $e^x$

Note: I would like to thank Alexander Gibson for pointing out that this proof only applies to the set of analytic functions.

Once again, let $f(x)$ be a function so that $f(x)=f'(x)$. Notice that this implies the following:

$f(x)=f'(x)=f''(x)=...=f^{(n)}(x)$

because

$f(x)=f'(x)\implies\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h\to 0} \frac{f'(x+h)-f'(x)}{h}\implies f'(x)=f''(x)$

and so forth. We can also say that

$f(a)=f'(a)=f''(a)=...=f^{(n)}(a)=C$

for some constant $C$. Use the theory of Taylor series to expand $f(x)$:

$f(x)=f(a)+f'(a)(x-a)+f''(a)\frac{(x-a)^2}{2!}+...+f^{(n)}(a)\frac{(x-a)^n}{n!}+...$

for $f(x)$ around any point $x=a$. Inputting our initial assumption into the Taylor expansion of $f(x)$ yields:

$f(x)=C+C(x)+C\frac{(x-a)^2}{2!}+...+C\frac{(x-a)^n}{n!}+...$

Divide both sides by $C$:

$\frac{f(x)}{C}=1+(x-a)+\frac{(x-a)^2}{2!}+...+\frac{(x-a)^n}{n!}+...$

Notice that the right hand side is the Taylor expansion of $e^x$ about a point $x=a$. We can conclude that

$\frac{f(x)}{C}=e^x$

$f(x)=Ce^x$

Note: It was okay to assume that the Taylor series converges, because it always does in a small interval about $x-a$.

Whew! That was a lot of theory. If you want to share your own proof of the unique self-differentiation property of $Ce^x$, then share it down below in the comments.