I've been reading this post. It says that $\sqrt[m]{x^n} = x^{n\frac 1m}=x^{\frac mn}=x$ if $m=n$. Let's take $x=-2$, and $m=n=2$. Now we have,
$\sqrt[2]{(-2)^2}=\sqrt[2]{4}=2$
But according to that answer we must have $\sqrt[2]{(-2)^2}=(-2)^{2/2}=-2 $ So I guess that formula $\sqrt[n]{a^m}=(a^m)^{\frac1n}=a^{m\cdot\frac1a}=a^{\frac mn}=a^{\frac1n\cdot m}=(\sqrt[n]a)^m$ is not true for all real numbers. Also the other answer suggest that the formula holds only when $n≥0,m>0,x≥0$.
So the question is what is the correct formula which hold for negative values of $x$ too and where can I find a good explanation and proof of it. I've gone through the wikipedea artical http://en.wikipedia.org/wiki/Exponentiation about it. It is very long and complex. I could not find a satisfactory explanation there.
You are correct in noticing that the principle value of $\sqrt[n]{x^n} \neq x$ when $x$ is negative and $n$ is even. In this instance, though, you can say that $\sqrt[n]{(x)^n} = |x|$