Proving a language is not regular using the pumping lemma Let
$$
L 
=\Big\{ \ 
ba^{2^k}b^{i_1}ab^{i_2}a\dots b^{i_k}a
\ \Big|\ 
k\ge 1,\ i_j\ge1 \ ,\ 1\le j \le k\ 
\Big\}\ .
$$
Using the pumping lemma prove that $L$ isn't regular.
The answer given to this is:

Assuming $L$ is regular, from the pumping lemma there is $n\in\mathbb{N}$ such that for $z=ba^{2^n}b^{i_1}ab^{i_2}a\dots b^{i_n}a$,
$z\in L$, $|z|\ge n$ can be expressed as $z=uvw$ where $u=ba^s (0\le s\le n-2)$, $v=a^t (1\le t\le n)$, and
$w=a^{2^n-s-t}b^{i_1}ab^{i_2}a...b^{i_k}a$. Choosing $i=2$ we get
$$
\begin{aligned}
uv^iw
&=b\ a^sa^{2t}a^{2^n-s-t}\ b^{i_1}ab^{i_2}a\dots b^{i_k}a \\
&=b\ a^{2^n+t}\ b^{i_1}ab^{i_2}a\dots b^{i_k}a
\end{aligned}
$$ but $n>1$ so $$2^n < 2^n + t \le 2^n + n < 2^n + 2^n = 2\cdot 2^n =2^{n+1}$$ a contradiction therefore $L$ is not
regular.

I don't understand why the last inequality leads to a contradiction
 A: Note that a string $x\in L$ if and only if it takes the form $ba^{2^k}b\text{(stuff)}$. In particular, the number of $a$'s in the first "block" must be a power of $2$. Here, you see that the number of $a$'s in the first block lies strictly between $2^n$ and $2^{n+1}$ and thus cannot be a power of $2$, yet $x$ is in $L$, a contradiction.
A: If the language were regular, the pumped word
$$
uv^2w=ba^{2^n+t}b^{i_1}ab^{i_2}a\dots b^{i_k}
$$
would have to be in $\ L\ $, which would require $\ 2^n+t=2^k\ $, by the definition of $\ L\ $.  But since $\ 2^n<2^n+t<2^{n+1}\ $, it is impossible for $\ 2^n+t\ $ to be any integral power of $\ 2\ $.
A: This answer is not exactly answering the question about why the given argument is working. Because the argument makes a contradiction work for a lot of strings of a general shape involving variables $i_1,i_2,\dots,i_k$. And the contradiction is not really transparent in such a not needed generality. Instead, it would be simpler to take one special word in $L$, show a contradiction for this one word. This is done in the answer, the argument being related with the one in the question.

We note first the following special requirement for an element of $L$. Each word $w\in L$ has a specific (let's call it) "key" $k$ among $1,2,3,\dots$ being explicitly of the shape
$$
\begin{aligned}
w &= ba^2b^{i_1}a    &&\text{ with key $k=1$ or }
\\
w &= ba^4b^{i_1}ab^{i_2}a    &&\text{ with key $k=2$ or }
\\
w &= ba^8b^{i_1}ab^{i_2}ab^{i_3}a    &&\text{ with key $k=3$ or }
\\
&\vdots
\\
w &= ba^{2^k}b^{i_1}ab^{i_2}a\dots ab^{i_k}a
&&\text{ with key $k$ or }
\\
\vdots
\end{aligned}
$$
and apply now the pumping lemma, assuming that $L$ is regular.
(We try to get a contraction. Above, the $b$-powers $i,j,s,i_1,i_2,\dots,i_k$ are each $\ge1$. Note that the "first subword $a$-power" is at least $a^2$.)
Then there exists some pumping length $p$, so that for each word $w\in L$ of length $|w|\ge p$ can be split as $w=xyz$ with $y$ non-empty, $xy$ has length $\le p$, and all words $xz$, $xyz=w$, $xyyz$, $xyyyz$, ... are also in $L$.
Let us consider the first / minimal $n$ so that $2^n>p$. Consider the following special element in $L$:
$$
w = b\ a^{2^n}\ \underbrace{b\ a\ b\ a\ b\ a\ \dots\ a\ b\ a}_{n\text{ occurrences of }b}\ .
$$
Then $|w|=1+2^n+2n>p$. This word has the "key" equal to $n$.
Write this $w$ as $w=xyz$, as insured by the pumping lemma, and from $|xy|\le p<2^n$ we obtain $xy$ as a left substring of $ba^{2^n}$.

*

*The case of $x$ empty is immediately excluded, since then $y$ is of the shape $ba^s$, and then a word of the shape $xyyyy\dots yz$ has the "key" determined from its left $xy$ part, but can contain then more $b$-occurrences then the "key" allows, from the $yyy\dots y$ part, contradiction.


*The case of $x$ not empty makes $x$ of the shape $x=ba^s$ with $s\ge 0$ and so $y=a^t$ with $t\ge 1$. Then every word
$$w_q:=x\underbrace{yyy\dots y}_{q\text{ times}}z$$ has the same amount $(n+1)$ of $b$'s, the same one as $xyz=w_1=w$, it determines thus a fixed "key" equal to $n$, but then the $y^q=yyy\dots y$ part in $w_q$ gets length bigger $2^n$ already for $q=2$, contradiction. (Or delete $y$, i.e. $q=0$ to get a shorter length and an other contradiction.)
$\square$
