$\sqrt[5]{2^4\sqrt[3]{16}} = ? $ $$\sqrt[5]{2^4\sqrt[3]{16}} = ? $$
Rewriting this and we have
$$\sqrt[5\cdot 3]{2^{4\cdot 3}4}$$
$$\sqrt[15]{2^{12}2^2}$$
Finally we get
$$\sqrt[15]{2^{12}2^2} = \sqrt[15]{2^{14}} = 2^{\frac{15}{14}}$$
Am I right?
 A: \begin{align}
\sqrt[5]{2^4 \sqrt[3]{16}}
&=\sqrt[5]{2^4 \sqrt[3]{2^4}}
=\sqrt[5]{2^4 \sqrt[3]{2^3 \cdot 2}}
=\sqrt[5]{2^4 \cdot 2 \sqrt[3]{2}}\\
&=\sqrt[5]{2^5 \sqrt[3]{2}}
=2 \sqrt[5]{\sqrt[3]{2}}
=2 \cdot \sqrt[15]{2}
=2 \cdot 2^{1/15}\\
&=2^{16/15}.
\end{align}
A: $$2^4\sqrt[3]{16}=2^4\cdot2^{4/3}=2^{16/3}\implies\sqrt[5]{2^4\sqrt{16}}=\left(2^{16/3}\right)^{1/5}=2^{16/15}$$
so no, you are not right...but almost.
A: In your first reformulation, you should get $\sqrt[15]{2^{4 \cdot 3} \cdot 16}$ and in the last part note that $\sqrt[a]{c^b} = c^{\frac{b}{a}}$ and not $c^{\frac{a}{b}}$.
A: $\sqrt[5]{2^4\sqrt[3]{16}} =$
$\sqrt[5\cdot 3]{2^{4\cdot 3}\color{blue}{16}}=$
$\sqrt[15]{2^{12}2^{\color{blue}{4}}}=$
$\sqrt[15]{2^{12}2^{\color{blue}{4}}} = \sqrt[15]{2^{\color{blue}{16}}} = 2^{\frac{\not 1\not 5\color{red}{16}}{\not{\color{blue}{1\not 6}}\color{red}{15}}}$
But in my opinion your method seems a little scatter-shot and undirected.  In particular $\sqrt[k]{b} = \sqrt[mk]{b^m}$ rubs me the wrong way.  It's not wrong per se, but is seems that we are going in the wrong direction and making things complicated rather than simpler.  And I fear extraneous roots and sign errors.  (Ex:  $\sqrt[5]{(-1)^3} \ne \sqrt [5*2]{(-1)^{3*2}}$).
Although there is no universal right or wrong way to do things try to develop a more systematic approach.  I'd personal reduce to a common base, convert radicals to fractional exponents, and then just do the math:
$\sqrt[5]{2^4\sqrt[3]{16}}=$
$\sqrt[5]{2^4\sqrt[3]{2^4}} =$
$(2^4(2^4)^{\frac 13})^{\frac 15}=$
$2^{\frac 15(4 + 4*\frac 13)} =$
$2^{\frac{16}{15}}$.
Also, I suppose I should point out that:
$2^{\frac {16}{15}} = 2^{1  \frac 1{15}} = 2\sqrt[15]{2}$ 
which could be an acceptable answer.  As is $2\times 2^{\frac 1{15}}$.
Which of these three answers $2^{\frac {16}{15}}, 2\sqrt[15]{2}, 2\times 2^{\frac 1{15}}$ it the correct one?  Well, none are, or they all are.
I personally prefer $2\sqrt[15]{2}$ as it.... well, I get a better sense of what makes the number.  I would prefer $2\times 2^{\frac 1{15}}$ as I'd general like to conform radicals and exponents, but in this case requiring the $\times$ sign bugs me.
But this is purely subjective.
A: You wrote $$\sqrt[5]{2^4\sqrt[3]{16}}=\sqrt[5\cdot 3]{2^{4\cdot 3}4}$$ 
which is not true because
$$\sqrt[3]{16}=16^\frac{1}{3}$$
That there is the third root of $16$, not $2^{4\cdot 3}$. Also you can not write $$\sqrt[5]{something\cdot \sqrt[3]{something}}=\sqrt[5\cdot 3]{something\cdot something}$$
The way to do it is:
$$\sqrt[5]{2^4\sqrt[3]{16}}=\sqrt[5]{2^4\cdot16^\frac{1}{3}}=\sqrt[5]{2^4\cdot2^\frac{4}{3}}=\sqrt[5]{2^\frac{16}{3}}=2^\frac{\frac{16}{3}}{5}=2^\frac{16}{15}$$
A: $$\sqrt[5]{2^4\sqrt[3]{16}} = (2^4)^{1/5}(16^{1/3})^{1/5}=2^{4/5}(2^4)^{1/15}=2^{4/5}2^{4/15}=2^{16/15}$$
