Proof involving functional equation

I'm trying to prove that if $$f(x+n)=f(x)f(n)$$ for all $x\in \Bbb R$ and $n \in \Bbb N$, then it also holds for $x,n \in \Bbb R$. One "argument" I came up with was regarding the symmetry. There's no reason why one should be constrained to the integers, while the other can be any real number, but that's not a proper argument.

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Another thing I thought of is this: If we set $n=1$ then we get $$\begin{align} f(x+1) &= f(1)f(x) \tag{1} \end{align}$$ which is true for all $x \in \Bbb R$, now if we instead set $x=1$ then we get $f(n+1)=f(n)f(1)$ which must also be true for $n \in \Bbb R$ because $(1)$ is. What keeps me from being satisfied is that $n\in \Bbb R$ under the assumption that $x=1$.

Is my reasoning valid or not?

Edit: Forgot an important bit of information: $f$ is a continuous function.

Answer

Your proof is not valid; you are noticing that if $x$ is an integer then $n$ could vary anywhere in the reals, but this is just stating that the relation $f(x+n)=f(x)f(n)$ has a symmetry where we swap $x$ and $n$ and doesn't tell us anything about whether $f(x+y)=f(x)f(y)$ when both are real.

More directly, the statement you're trying to prove is false, so obviously your proof of it is not valid. Let $f$ be the function
$$f(x)=a^{x+g(x)}$$
where $g(x)$ is a function such that $g(x+1)=g(x)$ and $g(0)=0$, and clearly, for any integer $n$, it holds that $g(x+n)=g(x)$ and $g(n)=0$. Then, for integer $n$ and real $x$, it olds that
$$f(x+n)=a^{x+n+g(x+n)}$$

but we can write $x+n+g(x+n)=(x+g(x))+(n+g(n))$ so we have

$$f(x+n)=a^{x+g(x)}a^{n+g(n)}=f(x)f(n).$$
However, it's easy to choose reals for which $f(x+y)=f(x)f(y)$ does not hold; for instance, choosing $x=y=\frac{1}2$ and letting $k=g(\frac{1}2)+\frac{1}2$, we get
$$f(1)=f\left(\frac{1}2\right)f\left(\frac{1}2\right)$$
$$a = a^k\cdot a^k = a^{2k}$$
which does not hold if $a\neq 1$ and $g(\frac{1}2)\neq 0$.