Take $2^{1/n}$. Since $1/n$ can be simplified as $n^{-1}$, the original term can become $2^{n^{-1}}$. The exponents can then be multiplied to result in $2^{-n}$ which is $1/(2^n)$. However it is obvious that
$2^{1/n} \ne 1/(2^n)$
because the first item approaches 1 as $n \rightarrow \infty$ but the second item approaches 0. What is wrong with the logic I used to simplify the first into the second that causes this discrepancy?
On
The exponents cannot be multiplied, as the first exponent is raised to the second. You are right that $(x^a)^b=x^{ab}$, but you do not have $x^{a^b}=x^{ab}$, as $a^b \neq ab$.
On
The order in which you do the operations is vital here. Unlike addition and multiplication, exponentiation is not associative: $$a^{(b^{\large c})} \neq \left(a^b\right)^c$$ for "most" values of $a$, $b$, and $c$.
It is true that $2^{1/n} = 2^{n^{\large -1}}$, but that's because $2^{n^{\large -1}}$ means $2^{(n^{\large -1})}$. As you saw for yourself, $2^{(n^{\large -1})}$ is not the same thing at all as $\left(2^n\right)^{-1}$.
Exponents multiply in the sense that $$(a^b)^c = a^{bc}.$$ They do not multiply in the sense that $$a^{(b^c)} \ne a^{bc}.$$