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> Suppose that the function $f:\mathbf R\to\mathbf R$ satisfies $f(n)=e^n$, where $n\in \mathbf Z$, and $f(x)f(y)=f(x+y)$ for all $x,y\in\mathbf R$. Must it be the case that $f=\exp$?
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**What I have tried so far:**

If we can prove that $\lim\_{h\to0}f(h)=1$, then it follows that $$\lim\_{h\to0}f(a+h)=\lim\_{h\to0}f(a)f(h)=f(a)\lim\_{h\to0}f(h)=f(a) \, ,$$
and so $f$ is continuous. With the extra hypothesis of $f$ being continuous, it can then be shown that $f=\exp$. So if the statement is true, then the problem reduces to proving that $\lim\_{h\to0}f(h)=1$. If the statement is false, then there is a function $f$ satisfying the hypotheses of the question, but not satisfying $\lim\_{h\to0}f(h)=1$.