Proving all exponential properties from $b^{x+k}=b^x\cdot b^k$ Let's say that there is a function $f(x)$ which satisfies the following property:
$$f(x+k)=f(x)\space{}\cdot\space{}f(k)$$
In addition, $f(1)=b\gt1$. I am trying to prove the following property without explicitly relying on the fact that $f(x)=b^x$ (circular reasoning) or requiring that $n\in\mathbb{Z}^+$:
$$[f(x)]^n=f(nx)$$
The only way I've been able to prove it is thinking about $nx$ as being $x+x+x+x+\space{}...\space{}$ n times, and then using the assumed property of $f(x)$ to break this up as a product $n$ times which can then be written as an exponent by definition. However, this argument only really makes sense if $n$ is a natural number, but this property should hold for all $n\in\mathbb{R}$. Is there a way of extending or altering this argument so that it is still sensible for any real number value of $n$?
 A: Define $f(x+y)=f(x)\cdot f(y)$
by induction we prove that $[f(x)]^n=f(nx)$
Base case $[f(x)]^1=f(1\cdot x)=1\cdot f(x)=f(x)$
now suppose the property is true for $n$ that is $[f(x)]^n=f(nx)$
and let's prove it for $(n+1)$
$[f(x)]^{n+1}=[f(x)]^n f(x)=f(nx)f(x)=f(nx+x)=f(n+1)x)$
Proved.
A: *

*First let examine $f(0)$:

$f(0)=f(0+0)=f(0)f(0)\iff f(0)(f(0)-1)=0$ therefore $f(0)\in\{0,1\}$
If $f(0)=0$ then $f(x)=f(x+0)=f(x)f(0)=0$ and $f$ is the null function, which is not very interesting.
From now on let set $f(0)=1$ and $f(1)=b$.

*

*Let examine $f(n)$:

$f(n+1)=f(n)f(1)=bf(n)=b^2f(n-1)=\cdots=b^{n+1}f(0)=b^{n+1}$
So by induction (base case $f(0)$ and $f(1)$ verified, then $f(n)=b^n,\ \forall n\in\mathbb N$

*

*Let examine $f(-x)$:

$1=f(0)=f(x-x)=f(x)f(-x)\implies f(-x)=\dfrac 1{f(x)}$
In particular $f(-n)=\dfrac 1{f(n)}=\dfrac 1{b^n}=b^{-n}$ and we have extended to all $\mathbb Z$.

*

*Let examine $f(\frac pq)$:

By the same induction used for $f(n)$ we have $f(nx)=f(x)^n$ for $n$ natural, and use $f(-nx)=\frac 1{f(nx)}$ to extend to all integers.
In particular $b=f(1)=f(\frac qq)=f(\frac 1q)^q$ therefore $f(\frac 1q)=b^{\frac 1q}$
And $f(\frac pq)=f(\frac 1q)^p=b^{\frac pq}$.
Note in the same way we have $f(\frac pqx)=f(x)^{\frac pq}$

*

*Now we use continuity of $f$:

If we don't assume continuity we are stuck with the rationals. So since $\mathbb Q$ is dense in $\mathbb R$, we can extend $f$ to the reals and we have $f(x)=b^x$ and the formula $f(xy)=f(x)^y$ by continuity too from the last note.
