$f ≠ 0$ everywhere if $f : \mathbb{R}\to\mathbb{R}$ and $f = f'$ and $f(0) = 1$?

Let $f:\mathbb R\rightarrow \mathbb R$ be a differentiable function, and suppose $f=f'$ and $f(0)=1$. Then prove $f(x)\neq 0$ for all $x\in \mathbb R$

The way I solve this is kind of strange.

I first suppose there is a closed interval $[0, a]$ on the real line. Since $f$ is differentiable, there exists a $x_0\in[0, a] $ such that $\frac{f(a)-f(0)}{a}=f'(x_0)=f(x_0)$. Thus when $f(a)=1$, $f(x_0)=0$. Then I just let $f(a)=1$ and try to find some contradictions. Since $f(a)=f(0)=1$, by Rolle thorem, there should exist a $x_1\in[0, a]$ such that $f'(x_1)=f(x_1)=0$ and this $f(x_1)=0$ is supposed to be the maximum or minimum on the interval $[0,a]$. Then apply MVT again on the interval $[0, x_0]\implies\frac{f(x_0)-f(0)}{x_0}=\frac{-1}{x_0}=f'(x_2)=f'(x_2)\implies-1=x_0f(x_2)\implies f(x_2)<0$ for a point $x_2\in[0, x_0]$. The existence of $x_2$ make sure that $f(x_1)=0$ is not the minimum on $[0, a]$ and since $0<f(0)=f(a)=1\implies f(x_1)$ is not the maximum on the $[0, a]$. Also if there exists other points $\beta$, for example, and $f(\beta)$ is the maximum of $[0,a]$ this implies that $f'(\beta)=0=f(\beta)$ the contradiction remains. Thus $f(a)\neq 1\implies f(x_0)\neq 0$. Since $a$ is an arbitrary real number, this means $f$ has no zero point on $[0,\infty]$. By the similar idea $f$ doesn't have zero point on $[-\infty, 0].

Is this a correct idea? And any shorter version of proof？ Thanks in advance!

Let $g(x) = f(x)e^{-x}$. Then, $g$ is a differentiable function on $\mathbb R$. By the product rule, $g'(x) = 0$. So, $g$ is constant. Since $f(0)=1$, we have $g\equiv 1$. This proves $$f(x)=e^x$$ so that $f$ has no zeros.

There is a shorter proof if you know the Picard-Lindelöf theorem. Assume that there exists $x_0\in \mathbb{R}$ such that $f(x_0)=0$. Then $f$ solves the ODE $$ \begin{cases} f'&=f \\ f(x_0)&=0. \end{cases} $$ Note that $g\equiv 0$ also solves the ODE above, but by the uniqueness part of Picard-Lindelöf we would have $f=g$ which contradicts $f(0)=1$.

Similar to the work of yearning4pi on this topic:

With

$f'(x) = f(x), \; f(0) = 1, \tag 1$

we consider the function $e^{-x}f(x)$; we have

$(e^{-x}f(x))' = -e^{-x}f(x) + e^{-x}f'(x) = e^{-x}(-f(x) + f'(x)) = 0 \tag 2$

in light of (1); thus

$e^{-x}f(x) = c, \; \text{a constant}, \tag 3$

whence

$f(x) = ce^x; \tag 4$

then

$c = ce^0 = f(0) = 1, \tag 5$

so that

$f(x) = e^x; \tag 6$

clearly

$e^x = \displaystyle \sum_0^\infty \dfrac{x^n}{n!} > 0 \; \text{for} \; x \ge 0, \tag 7$

and if $x < 0$, $-x > 0$, so

$e^{-x} > 0; \tag 8$

but

$e^{-x}e^x = e^{-x + x} = e^0 = 1 \Longrightarrow e^x = \dfrac{1}{e^{-x}} > 0; \tag 9$

thus

$\forall x \in \Bbb R, \; f(x) = e^x > 0. \tag{10}$

And then, of course, we always have Picard-Lindeloef as invoked by our colleague Severin Schraven.