Irregularity of language of words whose length is of power of 2 How to show that the language containing the words whose length is a power of $2$, $L=\{w\mid|w|=2^i\}$, isn't regular using the pumping lemma?
The pumping lemma says that

Let be M a regular language. Then it exists a number $p&gt;0$ such that for each word $w\in L$, such that $|w|\ge p$, it exists $x,y,z$ such that :

$w=xyz$
$|y|&gt;0$
$|xy|&lt;p$
For each $i\ge 0$ we have $xy^iz\in L$.


(I don't perfectly understand this lemma...). I have great difficulties to understand the examples which I can provide you in French here.
My question is different from How to prove that a language is not regular? as far as all answers are based on $L=\{w\mid|w|=a^pb^q\}$ scheme, mine only have one element and is playing on the size rather than the scheme.
My attempt
We take $w=a^{2^n}$. Therefore we have $w\in L$ and $|w|\ge p$. We need a partition $w=xyz$ (to fill in condition 1.) such that $|xy|&lt;p$ (condition 3) and $|y|&gt;0$ (condition 2).

Let's assume $L$ is regular.
I thought about taking $|w|&gt;n$
$x = aaaa....$ , $y = aaaa...$,
$|x|=s$, $|y|=k$. We need to consider ALL the options, that is all
the possible $s,k$ such that $s≥0$,$k≥1$ and $s+k≤n$.

Let's take $i=0$, then $xy^iz=xz=a^{n-k}a^n\not\in L$ whatever may $s,k$ be and since $k\ge 1$, $L$ isn't regular at all (but why) and we reach a contradiction.
I have some difficulties understanding the conclusion.
The strategy for proving a language non-regular is a proof by contradiction: 

We assume that $L$ is regular. This then means that there exists a $p &gt; 0$ such that every string $w \in L$ where $|w| \geq p$ can be partitioned as $s = xyz$ such that this partition satisfies conditions 1, 2 and 3. But now it is enough to find some $w$ where $|w| \geq p$ such that no partition can satisfy all three conditions simultaneously.

In the concrete case, you have not told us what alphabet $\Sigma$ we are considering. I assume that $\Sigma = \{ a,b\}$, but this is not important. 

So we assume that $L$ is regular, and therefore there must exist a $p &gt; 0$ such that every string $w \in L$ where $|w| \geq p$ can be partitioned as $s = xyz$ such that this partition satisfies conditions 1, 2 and 3. 

But now consider $w = a^{2^p}$. Clearly, $w \in L$. Moreover, we have that $|w| \geq p$, so there should be some partition of $w = xyz$ that satisfies conditions 1, 2 and 3.

If the partition must satisfy condition $3$ then the first two parts of the partition, that is, $xy$, must have total length less than $p$. Therefore, $|y| \leq p$. If the partition must also satisfy condition $2$, then $y$ cannot be the empty string, so $y = a^k$ for some $k$ where $0 &lt; k &lt; p$.

But can such a partition then also satisfy condition $1$? No. It is enough to find an $i \geq 0$ such that $xy^iz \notin L$ to see this. Take $xy^2z$. This string has length $2^p + k$, and since $k &lt; p$, we know that $2^p + k$ cannot be a power of $2$. So $xy^2z \notin L$.

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In other words: The appropriate strategy is to try to satisfy the conditions one by one. Usually, we first try to satisfy condition $3$. This will tell us where $y$ can appear (among the first $p$ symbols) and how long it can be. Given that, we then try to also satisfy condition $2$, and this usually just provides with the useful information that the string $y$ that can be repeated, is non-empty. Finally, we show that we can "pump our way out" of the language with this partition, thereby violating condition $1$.