How to define the exponential function without calculus? For fun, I would like to define the complex exponential function from these two properties:


$\exp(0) = 1$
$\exp(z + w) = \exp(z) \exp(w)$


From here, I would like to find a way to compute values of $\exp(z)$, or at least to compute $\exp(1)$. 

So far, I found only two ways:


Noting that $\exp'(z) = \exp(z)$ and solving the differential equation, which leads to $\int \frac{\exp'(z)}{\exp(z)} dz = \log(\exp(z)) + C = z$.
Noting that $\exp'(z) = \exp(z)$, computing its Taylor series and checking that what I get is an entire function.


The first approach is simply wrong because it involves logarithms, which I have not defined yet. The second approach looks much better. I haven't tried, but I guess I can find a way to manipulate the Taylor series to obtain the limit definition of $e$ and conclude that $\exp(1) = e$, which is my aim.

However, I'm struggling to find another way that does not involve differentiation or limits in general. I would be happy to find a way to say $\exp(1) = e$ without calculus. I think that the irrational nature of $e$ forces me to use limits -- am I right?
You won't be able to do derive

$exp(1)=e$ 

from your definition, since your definition works for the exponential function for any base $b$:

$b^0=1$

and 

$b^{x+z}=b^x\cdot b^z$

are true for any $b$!
(An echo of previous answers)

If
$f(x+y)
= f(x)f(y)
$
and $f$ is differentiable,
then
$f(0) = 1$
(set $y=0$),
so
$\lim_{x \to 0} f(x)
= 1
$.

Switching to standard calculus variables,

$\begin{array}\\
f(x+h)-f(x)
&amp;=f(x)f(h)-f(x)\\
&amp;=f(x)(f(h)-1)\\
&amp;=f(x)(f(h)-f(0))\\
\text{so}\\
\dfrac{f(x+h)-f(x)}{h}
&amp;=f(x)\dfrac{f(h)-f(0)}{h}\\
\end{array}
$

Letting
$h \to 0$,
$f'(x) = f(x)f'(0)$.

$f(x)=a^x$ satisfies
$f(x+y)=f(x)f(y)$.
$a=e$ is special
because that makes
$f'(0) = 1$.

In general,
$(a^x)' = a^x\ln(a)
$.