Prove that $\sqrt[kn]{a^{km}}=\sqrt[n]{a^m}$

Question

$a>0$.

I'm trying to prove that $\sqrt[kn]{a^{km}}=\sqrt[n]{a^ m}$.

Here's what I tried to do, but I think it's useless.

$\sqrt[kn]{a^{km}}=\sqrt[kn]{(a^m)^k} = ((a^m)^k)^{1/kn} = \Bigr(\bigr((a^m)^k\bigr)^{1/k}\Bigr)^{1/n} = (a^{m/n})^k$.

Answer 1

By your work it should be $$\sqrt[kn]{a^{km}}=\sqrt[kn]{(a^m)^k} = ((a^m)^k)^{\frac{1}{kn}}=(a^m)^{k\cdot\frac{1}{kn}}=a^{\frac{m}{n}}=\sqrt[n]{a^m}$$

Answer 2

By definition

$$ \sqrt[kn]{a^{km}}=a^{\left(km\right)/\left(kn\right)}=a^{\left(m/n\right)}=\sqrt[n]{a^{m}} $$