$f ≠ 0$ 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!
 A: 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.
A: 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 $f(0)=1$.
A: 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.
A: Certainly the shortest solution involves $e^x$. However, depending on one's definitions, such a solution may very well involve a circular argument: $e^x$ is often defined directly as the unique function $f$ such that $f'(x) = f(x)$ for all $x$ and $f(0) = 1$. Let us therefore try to come up with a direct, elementary argument which does not involve $e^x$.
We first show that there are no solutions of $f(x) = 0$ for $x \geq 0$.
Observation. If $f(a) > 0$, then $f(b) > 0$ for $b > a$.
Proof of observation. Let $b$ be the infimum of all $x > a$ such that $f(x) = 0$ (or if you wish, $f(x) \leq 0$). Because the function $f$ is continuous, in fact $f(b) = 0$. In other words, $b$ is in fact the smallest $x > a$ such that $f(x) = 0$. Clearly $b > a$. Because $f(a) > 0$ and $f(b) = 0$, there must be some $c \in (a, b)$ where $f'(c) < 0$. But this contradicts the fact that $b$ is the smallest $x > a$ such that $f(x) = 0$, since $c < b$ and $f(c) = f'(c) < 0$.
It remains to show that there are no solutions of $f(x) = 0$ for $x < 0$. The set of solutions of $f(x) = 0$ is bounded from above by $0$, therefore it has a supremum $c$ if it is non-empty. Again, because the function $f$ is continuous, in fact $f(c) = 0$. In other words, $c$ is the largest $x$ such that $f(x) = 0$. Clearly $c < 0$.
The following can also be phrased in terms of derivatives, but it is perhaps easier to visualize if we talk about parallel lines instead.
Observation. Suppose that the graph of $f$ is tangent to a line $L$ at some point $\langle x_0,f(x_0) \rangle$ where $x_0 > c$. Then the graph of $f$ lies strictly above $L$ on the interval $(x_0, \infty)$ and it lies strictly below $L$ on the interval $[c, x_0)$.
Proof of observation. Taking $g(x) := f(x) - L(x)$, we need to show that $g(x) > 0$ for $x \in (x_0, \infty)$ and $g(x) < 0$ for $x \in (c, x_0)$. This follows from the fact that $g(x_0) = 0$, $g'(x_0) = 0$, and $g''$ is strictly positive on $(c, \infty)$.
With this last observation in hand, we simply apply the Mean Value Theorem to the interval $[c,0]$. This tells us that there is some $x_0 \in (c,0)$ such that $f$ is parallel to the line $L$ connecting the points $\langle c,0 \rangle$ and $\langle 0, 1 \rangle$. In other words, the graph of $f$ is tangent to a line $L'$ parallel to $L$ at $\langle x_0, f(x_0) \rangle$. Now if $L' = L$ or $L'$ lies above $L$, then the above observation tells us that we cannot have $f(0) = 1$. On the other hand, if $L'$ lies below $L$, then the observation tells us that we cannot have $f(c) = 0$. In either case, we obtain a contradiction.
A: Now I think the easiest way of dealing with this problem is just to solve the Cauchy problem $$\begin{align*}
\begin{cases}
y'=f(y)=y \\ \\ y(0)=1 
\end{cases}
\end{align*}$$ Here I use $y$ to denote $f$ . Since $y$ has a continuous partial derivative with respect to y in $\mathbb{R}$, so it satisfies the Lipschitz condition with respect to $y$. Thus, by existence and uniqueness theorem, this Cauchy problem has a unique solution, and since this ode is separable, we can solve it to get $$\begin{align*}
y=e^x
\end{align*}$$ which is the unique function that satisfies the required condition.
