nth root formulas

In the following formulas, $a$ is a nonnegative real numberand other letters positive integers.  For other formulas, seethe parent entry (http://planetmath.org/NthRoot).

1. $\sqrt[n]{0}=\mathrm{\hspace{0.33em}0}$,   $\sqrt[n]{1}=\mathrm{\hspace{0.33em}1}$

2. $\sqrt[1]{a}=a$

3. $\sqrt[m]{\sqrt[n]{a}}=\sqrt[mn]{a}=\sqrt[n]{\sqrt[m]{a}}$

4. $\sqrt[nk]{a^{mk}}=\sqrt[n]{a^m}$

5. $\sqrt[n]{a^m}={(\sqrt[n]{a})}^m$

6. $\sqrt[m]{a}\cdot \sqrt[n]{a}=\sqrt[mn]{a^{m+n}}$

Proof.  For proving, one uses the definition ofhttp://planetmath.org/node/754$\mathrm{n}$th root and the power (http://planetmath.org/GeneralAssociativity) laws.

$1^{\circ }$.  $0^n=0,1^n=1$

$2^{\circ }$.  $a^1=a$

$3^{\circ }$.  ${(\sqrt[m]{\sqrt[n]{a}})}^{mn}={({(\sqrt[m]{\sqrt[n]{a}})}^m)}^n={(\sqrt[n]{a})}^n=a$

$4^{\circ }$.  ${(\sqrt[n]{a^m})}^{nk}={({(\sqrt[n]{a^m})}^n)}^k={(a^m)}^k=a^{mk}$

$5^{\circ }$.  ${({(\sqrt[n]{a})}^m)}^n={({(\sqrt[n]{a})}^n)}^m=a^m$

$6^{\circ }$.  ${(\sqrt[m]{a}\cdot \sqrt[n]{a})}^{mn}={(\sqrt[m]{a})}^{mn}{(\sqrt[n]{a})}^{mn}={({(\sqrt[m]{a})}^m)}^n{({(\sqrt[n]{a})}^n)}^m=a^n{a}^m=a^{m+n}$