Solve for $f(x)=f'(x)$ without previous knowledge. 
Solve for $f(x)=f'(x)$ without previous knowledge.

I know it is obviously $f(x)=e^x$, but could you prove this without knowing $\frac d {dx}e^x=e^x$?
And does there exist a $g(x)=g'(x)$ but $g(x)\ne f(x)$?
 A: $$
f = \frac{df}{dx}
$$
$$
dx = \frac{df} f
$$
$$ x + \text{constant} = \log_e |f|
$$
$$
e^x \cdot\text{positive constant} = |f|
$$
$$
e^x \cdot\text{constant} = f.
$$
(One must check separately that $f=0$ is a solution.)
Suppose there were some other solution $g$ (maybe equal to $0$ at some points and not at others?  or whatever $\ldots$).  Then
$$
\underbrace{\frac d {dx}\ \frac{g(x)}{e^x} = \frac{e^x g'(x) - g(x)e^x}{e^{2x}}}_\text{quotient rule}\  \  \underbrace{ = \frac{e^xg(x) - e^x g(x)}{e^{2x}}}_{\text{since }g\,'\,=\,g} = 0,
$$
so $\dfrac{g(x)}{e^x} = \text{constant}$, and we can rule out other solutions.
A: Another way do to it, let $(a_k)_{k\in\mathbb{N}}\in\mathbb{R}^\mathbb{N}$ with positive radius of convergence $R$. Let define: $$\forall x\in ]-R,R[,f(x)=\sum_{k=0}^{+\infty}a_kx^k.$$
$f$ satisfies the equation if and only if: $$\forall k\in\mathbb{N}_{\geqslant 1},a_{k-1}=ka_k\Leftrightarrow a_k=\frac{a_{k-1}}{k}.$$
If $a_0=1$, then one has: $$f(x)=\sum_{k=0}^{+\infty}\frac{x^k}{k!}.$$
Using D'Alembert's criterion, one checks that $(a_k)_{k\in\mathbb{N}}$ has infinite radius of convergence.
A: $f'(x)=f(x)$ doesn't implies $f(x)=e^x$. But:

If $f(x)=f'(x)$ and $f(0)=1$ then $f(x)=e^x$.

Proof: It follows from the Taylor's Theorem that
$$f(x)=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$
for all $x\in\mathbb{R}$. Then,$$\begin{aligned}
f(x)f(y)&=\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\right)\left(1+\frac{y}{1!}+\frac{y^2}{2!}+\frac{y^3}{3!}+\cdots\right)\\\\
&=\left(1+\left(\frac{x}{1!}+\frac{y}{1!}\right)+\left(\frac{x^2}{2!}+\frac{x}{1!}\frac{y}{1!}+\frac{y^2}{2!}\right)+\left(\frac{x^3}{3!}+\frac{x^2}{2!}\frac{y}{1!}+\frac{x}{1!}\frac{y^2}{2!}+\frac{y^3}{3!}\right)+\cdots\right)\\\\
&=\left(1+\frac{(x+y)}{1!}+\frac{(x+y)^2}{2!}+\frac{(x+y)^3}{3!}+\cdots\right)\\\\
&=f(x+y)
\end{aligned}$$
for all $x,y\in\mathbb{R}$. Therefore,
$$f(n)=f(1+1+\cdots +1)=f(1)f(1)\cdots f(1)=f(1)^n,\qquad\forall\ n\in\mathbb{N}$$
which implies
$$f(r)=f(1)^r,\qquad\forall\ r\in\mathbb{Q}$$
which implies
$$f(x)=f(1)^x,\qquad\forall\ x\in\mathbb{R}$$
As $f(1)=\sum\frac{1}{k!}=e$, we get $f(x)=e^x$. $\square$
In short: If $f(x)=f'(x)$ and $f(0)=1$, then $f(x+y)=f(x)f(y)$ and $f(1)=e$. But there is only one function with these properties, namely, the function $e^x$. So, $f(x)=e^x$.
Remark 1: The same argument shows that $f(x)=f'(x)$ implies $f(x)=Ce^x$ for some constant $C\in\mathbb{R}$ (namely, $C=f(0)$).
Remark 2: The uniqueness follows from the proof.
Remark 3: A solution of this problem depends on what we know about $e$. In @MichaelHardy 's solution we need to know that $e=\log(1)$ where $\log(x)$ is defined as the integral of $1/t$ from $0$ to $x$; in this solution we need to know that
$$e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots$$
A: In relation to the second part of your question, there is no $g(x)$ as $e$ is defined at being a number such that $\frac{d}{dx}e^x = e^x$. clearly you can multiply by a constant as this won't affect the differentiation but that is the only adjustment you can make.
bacj=k to the first part;
$$y=e^x$$
$$\ln(y) = x$$
I will now use implicit differentiation
$$\frac{1}{y} \frac{dy}{dx} = 1$$
$$\frac{dy}{dx} = y = e^x$$
And we're done
