Solving $f(x)f(y) = f(x + y)$ I am a little lost trying to derive what form $f(x)$ must have if we know $f(x)f(y) = f(x + y)$ for real inputs $x, y$.
My attempt so far:
Set $y=0$ and we have $f(x)f(0) = f(x)$ meaning either $f(x) = 0$ or $f(0) = 1$. Not sure what to do with this.
What about setting $y=x$? Then $f(x)^2 = f(2x)$. Multiply both sides by $f(x)$ and then $f(x)^3 = f(2x)f(x) = f(2x + x) = f(3x)$ and so on, so $f(x)^n = f(nx)$ for some integer $n \geq 2$. But it's also true for $n=1$ because $f(x)^1 = f(1 \cdot x) = f(x)$ and it's also true for $n=0$ (if we assume $f(0) = 1$) since $f(x)^0 = f(0 \cdot x) = f(0) = 1$, so $f(x)^n = f(nx)$ holds for integer $n \geq 0$.
For $n > 0$: raise both sides to $1/n$ and we get
$f(x) = f(nx)^{1/n}$
I don't really know where I am going with this or if it's even the right track. Am I even allowed to do that in the first place? Am I supposed to be assuming $f(x)$ is real? Or complex? Or positive? Or something? Should I be assuming $x$ and $y$ are complex? I don't really know what assumptions to make exactly. I'm just trying to prove/show that this all implies $f(x)$ has some exponential form but pretending I don't know that yet.
Could use any corrections or a push in the right direction. 
 A: You're doing fine. So far you've managed to identify that either


*

*$f$ is everywhere zero, or

*$f(0) = 1$, in which case $f(nx) = f(x)^n$ for every positive integer $n$. 
You can probably also manage to show that for every positive integer $k$, you have $f(x/k) = f(x)^{1/k}$, and then combine these to conclude that for any rational number $r$, $f(rx) = f(x)^r$. 
A good next place to look is to say "let's say $f(1) = A$." Then we can work out $f(2), f(3), \ldots$ and $f(1/2), f(1/3), \ldots$, and maybe even $f(r)$ for every rational number $r$ with a little cleverness. 
But what about irrationals? To say anything useful there, I believe you need an added assumption like "$f$ is continuous". 
Post-comment additions
For things like this problem, it can be really helpful to write down everything in detail, rather than just as notes. You could, for instance, say this:
I'm studying the functional equation 
$$
f(x + y) = f(x)f(y), \tag{1}
$$
which I'll assume is defined for $x$ a real number, and that the values taken by $f$ are also real, i.e., that I have
$$
f: \Bbb R \to \Bbb R : x \mapsto f(x)
$$
Lemma 1: If $f(0) = 0$, then $f(x) = 0$ for all $x \in \Bbb R$. 
Proof: From equation 1, we have $f(x) = f(x + 0) = f(x) f(0) = f(x)\cdot 0 = 0.
Lemma 2: Assuming $c = f(0) \ne 0$, we have $f(0) = 1$. 
Proof: $f(0) = f(0 + 0) = f(0)^2$, so $c = c^2$, hence $c - c^2 = c(1-c) = 0$, when $c = 0$ or $c = 1$. We've assumed $c \ne 0$, hence $c = 1$. QED. 
Henceforth we'll assume $f(0) = 1$ and ignore the always-zero solution. 
Lemma 3: For any $x\in \Bbb R$, $f(2x) = f(x)^2; f(3x) = f(x)^3$. 
Proof: $f(2x) = f(x + x) = f(x) f(x)$ by equation 1. Similarly, breaking up $f(3x) = f(2x) + f(x)$ establishes the second claim. 
Lemma 4: For any positive integer $n$, $f(nx) = f(x)^n$. 
Proof, by induction: Let $P(m)$ be the statement that for the positive integer $m$, and for every real number $x$, $f(mx) = f(x)^m$. We know that for any real $x$, $f(1x) = f(x) = f(x)^1$, so $P(1)$ is true.  Suppose that for some integer $k$, we know $f(kx) = f(x)^k$ (this is our induction hypothesis $P(k)$). Then let's examine $f((k+1) x)$:
\begin{align}
f((k+1)x) 
&= f(kx + x) \\
&= f(kx)f(x) & \text{By equation 1} \\
&= f(x)^kf(x) & \text{By the induction hypothesis}\\ &= f(x)^{k+1}.
\end{align}
We see that $P(k)$ implies $P(k+1)$; combining this with the fact that $P(1)$ is true, we find (by induction) that $P(n)$ is true for all positive integers $n$.
...and you continue in this vein. It really helps to know what assumptions you're making in each step. 
A: Note that $f$ is nonnegative, since $f(x) = f(\frac{x}{2})^2$. If $f$ is positive, taking $\log$ of both sides gives $\log (f(x))+ \log (f(y)) = \log (f(x+y))$. Letting $g(x) = \log (f(x))$, we obtain the famous Cauchy’s functional equation $g(x+y)=g(x) + g(y)$, which is linear over the rationals, but which may be very ill-behaved over the reals (there exist solutions which are dense over $\mathbb{R}^2$, for example.) This will yield a large number of messy solutions for $f$, unless you specify some other constraint such as continuity, or boundedness as some point, etc. (various sufficient constraints that force $f$ to be linear are given here: https://en.m.wikipedia.org/wiki/Cauchy's_functional_equation)
A: A 'hint' is that $f(x+y)=f(x)f(y)$ is satisfied by $f(x)=a^x$ for all $a > 0$.
If $f(x) = 0$ for all $x \in \mathbb{R}$ then this is trivial, so assume it isn't.
Then $f(0)=1$, as you suggest. Let $a=f(1)$. Then you can show:


*

*$f(n) = a^n$ for all $n \in \mathbb{N}$ by induction.

*$f(n) = a^n$ for all $n \in \mathbb{Z}$ using the fact that $f(n + (-n)) = f(0) = 1$ for all $n \ge 0$.

*$f(x) = a^x$ for all $x \in \mathbb{Q}$ likewise.


Now if $f$ is required to be continuous, then this is all you need. [And if not, this would become very messy indeed.]
A: From $y=0$, we conclude that either $f(x)$ is identically zero (one solution), or else $f(0)=1$.  This gives one value for $f$ for all solutions other than $f(x)\equiv 0$.
From the second calculation, taking $x=1$, you have shown that $f(n)=f(1)^n$.  Further, we have $$f\left(\frac{n}{m}\right)^m=f\left(\frac{n}{m}\right)f\left(\frac{n}{m}\right)\cdots f\left(\frac{n}{m}\right)=f\left(\frac{n}{m}+\frac{n}{m}+\cdots+\frac{n}{m}\right)=f(n)=f(1)^n$$
Hence, $f\left(\frac{n}{m}\right)=f(1)^{n/m}$
In particular, knowing $f(1)$ determines $f(x)$ on all rational numbers.
However, this does not inform $f(x)$ on irrational numbers.  In fact, you could specify $f(\sqrt{2})$ to be arbitrary.  This would then determine $f(\sqrt{2}t)$ for all rational $t$.
Commonly one would have an assumption about $f(x)$ like continuity, that would forcibly relate the two.
