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I want to prove this : $(ab)^n = a^nb^n$ with a, b and n real numbers. I know how to prove this when n is an integer but not when n is a real number. I really don't know where to start to prove this. Can you help me ? Thank you.

Never
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    What definition of the exponential function are you using? – MSDG Dec 01 '18 at 23:18
  • I don't really understand the question but I think the definition of the exponential I'm using is : $a^b = a...a$ where a appears b times. – Never Dec 01 '18 at 23:23
  • That definition works if the exponent is an integer. To discuss properties of $a^n$ when $n$ is real one needs to define it first. There are multiple (ultimately equivalent) ways of doing this. – MSDG Dec 01 '18 at 23:25
  • Ok,I will look at this. Thank you – Never Dec 01 '18 at 23:27

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If you're using the usual interpretation of the exponent of real numbers, you can use the fact that $(ab)^n=e^{n\log(ab)}=e^{n(\log a+\log b)}=e^{n\log a+n\log b}=e^{n\log a}\cdot e^{n\log b}=a^n\cdot b^n$

Edit: The fact that $e^{x+y}=e^{x}e^{y}$ comes from the way we define the exponent. If it is as the solution to the differential equation $f(z)=f'(z); f(0)=1$ this is pretty easy, as then $$D(e^{c-z}e^z)=e^{c-z}e^z-e^{c-z}e^z=0$$ which gives us that $e^{c-z}e^z$ is constant and the result follows for $x,y$.

NL1992
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