What are necessary and sufficient conditions that $x^{yz} = (x^y)^z$. This is somewhat discussed here Question on roots of unity.
One example for when it fails is $e^{i2\pi r} \neq ((e^{i\pi})^2)^r = (-1^2)^r = 1^r = 1$
The question was motivated by some Fourier series proofs. Here is the logic I used, which gets the correct answer, but now I doubt whether the proof is correct: ($k$ is an integer)
$\exp(2\pi~i~k) = ((e^{i\pi})^2)^k = ((-1)^2)^k = 1^k = 1$
TL;DR: It depends on your definition but it's always true if $z$ is an integer.
One fundamental problem with answering this is that there is often an impression that there is a single sensible definition of what it means to raise one number to the power of another, however this is in general not actually true.
We can define exponentials by the properties we want them to have, the most obvious property (and the one I will use as a definition) is that $x^{y+z}=x^yx^z$.
(To keep this question simple I will restrict myself to complex exponentials but the ideas here can be used in many other contexts.)
Let us start by considering a map $f:\Bbb{Z}\rightarrow\Bbb{C}$. Let's suppose that $f(1)=x$ for some $x\in\Bbb{C}-\{0\}$.
Now if it is an exponential map we can see that $f(2)=f(1)f(1)=xx=x^2$ and similarly for all positive integers. Additionally $f(1)=f(1+0)=f(1)f(0)$ so we can conclude that $f(0)=1$. Lastly we can see $1=f(0)=f(n-n)=f(n)f(-n)$ so we conclude $f(-n)=f(n)^{-1}$.
This allows us to define a unique exponential map on the integers which has $f(1)=x$. This collection of maps has many nice properties like $x^{yz}=(x^y)^z$.
Problems start happening when we aren't only using integer powers.
Let us consider exponential maps $f:\Bbb{Q}\rightarrow\Bbb{C}$. And let us restrict our attention to the case that $f(1)=1$.
There is one obvious choice which is that $f(x)=1$ for all $x$. However there are other maps which satisfy our definition of an exponential map. One example is as follows.
For any rational number we can write it as $2^n\frac{a}{b}$ with $n$ an integer and so that $2$ does not divide $a,b$. Now we define $E_n$ as the $2^{-n}$th root of unity which is 'next around the unit circle from 1'.
Let $f(2^n\frac{a}{b})=(E_n)^a$. This function meets our definition of an exponential map and has $f(1)=1$ but is not identically $1$. This means that lots of nice exponential rules fail.
Things get worse if we look at exponential functions $f:\Bbb{R}\rightarrow\Bbb{C}$ (or $f:\Bbb{C}\rightarrow\Bbb{C}$). However it would be useful to be able to talk about arbitrary exponents of numbers. To allow this the function $\exp$ was defined. This gives a canonical value for exponents of $e$, however it doesn't tell us how to raise any other number to any power. If we want to do that we 'define' the following by appealing to the exponential rules.
$\exp(x)^y=\exp(xy)$
This now let's us 'define' the exponential of an arbitrary complex number by using logarithms. However there is a problem.
Consider $1=\exp(0)=\exp(2\pi i)$. This means we want to have $1^\frac{1}{2}=\exp(0)=1$ and $1^\frac{1}{2}=\exp(\pi i)=-1$. This is a contradiction. There are a couple of ways this can be fixed.
One way is to use so called principal branches and choose some particular value for the logarithm of every complex number and then always use that when calculating exponentials. For example the principal logarithm of 1 is 0 and so any power of 1 is defined as 1. The problem with doing this is that you have to lose the rule that $x^{yz}=(x^y)^z$.
One alternative is to define exponentials as multivalued maps. This does let you keep lots of exponential rules but now exponentials give a set of values instead of just one, which can cause problems or be undesirable in many situations.
This is the primary reason that the calculation you did doesn't work, it is because you are ignoring the complexities of the definition of the exponential.
Now that that has been covered I can give what I believe is the most accurate answer possible to your question.
$x^{yz}=(x^y)^z$ as long as $z$ is an integer (no matter what), alternatively if you define exponentials as multivalued maps it is always true. Any other cases will ultimately depend on the exact definition of exponentials that is being used.
(The proof that $x^{yz}=(x^y)^z$ if $z$ is a positive integer follows by writing $yz$ as the sum of $z$ copies of $y$ and then using the defining property of an exponential map. Negative integer values for $z$ then follow easily, along with the case $z=0$.)
