What is the difference between $1 - p^k$ and $(1-p)^k$? Let $p$ be the probability that an event happens, then $1-p$ is the probability of that event not happening. The probability of that event happening $k$ times (in a row), i.e. rolling the same number for $k$ dice rolls, or landing heads $k$ times (in a row), is $p^k$ and similarly, the probability of not doing so is $(1-p)^k$.
If we argue that the probability of an event not happening is $1-$ that probability, why isn't the probability of not $p^k$ not equal to $1-p^k$?
Is it, that $1-p^k$ is the probability of $E$ "not happening $k$ times", e.g. $k-1$ times or less and $(1-p)^k$ $E$ not happening exactly $k$ times?
 A: If you have two events, E occurring with probability $p$, and the complementary event NE occurring with probability $1 - p$, then


*

*$p^k$ is the probability that E occurs $k$ times in a row.

*$(1 - p)^k$ is the probability that NE occurs $k$ times in a row.

*$1 - p^k$ is the probability that "E occurs $k$ times in a row" does not happen.  This is not the same event as "NE occurs $k$ times in a row", it is the same as "NE occurs at least once in $k$ consecutive trials".

*$1 - (1 - p)^k$ is the probability that the event "NE occurs $k$ times in a row" does not happen.  This is not the same as "E occurs $k$ times in a  row", it is the same as "E occurs at least once in $k$ consecutive trials".


To put this in the context of a simple example, the events referenced above can be specialized to the following in the case of a coin flip


*

*Heads is flipped $k$ times in a row.

*Tails is flipped $k$ times in a row.

*"Heads is flipped $k$ times in a row" does not happen $\Leftrightarrow$ Tails is flipped at least once.

*"Tails is flipped $k$ times in a row" does not happen $\Leftrightarrow$ Heads is flipped at least once.


English can be difficult and squeamish when you attempt to use it precisely.  The phrase "Heads is not flipped $k$ times in a row" is ambiguous, so I prefer to avoid it.
Here's another example of the ambiguity of language: another answer to the question in your title, if taken literally, is
$$(1 - p)^k - (1 - p^k) = \sum_{i = 0}^k (-1)^i {n \choose i} p^i - 1 + p^k = ((-1)^k - 1)p^k + \sum_{i = 1}^k {n \choose i} p^i $$
which I don't believe was what you were after.
A: If $E$ is a probable event in a trial with constant probability of appearing  $p$, and we have a series of $k$ independent trials, then
$p^k$ is the probability that $E$ will happen in all the trials  
$(1-p)^k$ is the probability that $E$ will not appear at all in any of the $k$ trials  
$1-p^k$ is the probability that $E$ will not appear in all the trials - but it may appear in $1,...,k-1$ of them.
Realize that you have two different levels of "events, trials and probabilities" here : Looking at the whole $k$-trials setup, there is no event $E$: instead the events here are ordered sets of cardinality $k$, like $\{E,E,...,E\}$ (that has probability $p^k$),
$\{E^c,E^c,...,E^c\}$ (that has probability $(1-p)^k$ or
$\{E^c,E,...,E\}$ (that has probability $(1-p)p^{k-1}$etc. 
So 
$$1-p^k = 1- \text{Prob}(\{E,E,...,E\}) \neq \text{Prob}(\{E^c,E^c,...,E^c\}) = (1-p)^k$$
evidently, since there are many more possible events here than just $\{E,E,...,E\}$ and $\{E^c,E^c,...,E^c\}$.
