$L_1$ is not regular.
Assume it was. Consider a recognizing DFA with $k_0$ states. On one hand it will recognize $a^{k_0}\, b^{k_0+1}$. Operating on any input with prefix $a^{k_1}\; \text{for some}\; k_1 \leq k_0$, it will visit a state twice. Therefore by the pumping lemma, any $a^{q*k_1}\,b^{k_0+1}, q \in \mathbb{N^+}$ will be recognized, notably $a^{(\lceil\frac{k_0}{k_1}\rceil + 1)* k_1}\, b^{k_0+1}$, a contradiction.
The hard part about these kinds of arguments is getting the quantifiers right. The pumping lemma says:
If $L$ is regular, then there exists a $p \ge 1$, such that for all $w \in L$ with $|w| \ge p$, there exists a "splitting" $w = xyz$, such that for all $i \ge 0$, $xy^iz \in L$.
You can think of this as an adversarial game. You pick a $p$, your opponent picks a $w$, you pick the $xyz$, they pick the $i$. If you can show it's in $L$, you win. So, the pumping lemma can be thought of as: if $L$ is regular, you can always win this game.
Therefore, if you lose, the language is not regular. So let's switch the roles; you are now the aggressor. Your opponent picks a $p$, you pick a $w$, they pick a splitting, you pick an $i$, and if you can always win, then $L$ is not regular. This can be phrased more mathematically: (note that the quantifiers are now switched!)
Let $L$ be a language. If for all $p \ge 1$, there exists a $w \in L$ with $|w| \ge p$, such that for all "splittings" $w = xyz$, there exists an $i \ge 0$ such that $xy^iz \notin L$, then $L$ is not regular.
So when proving languages are not regular, you don't get to pick how it splits!
What we have to do is pick our $w$ very carefully. I'll phrase this as a game first, then as a proof.
Let your opponent pick $p$. We'll choose $w = a^pbba^p$. No matter how our opponent splits the string, the conditions of the game/lemma require that $|xy| \le p$. So $xy$ consists only of $a$s, and cannot contain the $b$s. We choose $i = 0$, which will cause the $a$s to be asymmetric, and $xz \notin L$.
Assume $L$ is regular, for the sake of contradiction. Then $L$ satisfies the pumping lemma, so let $p$ be the pumping length. Consider $w = a^pbba^p$, where $a, b \in \Sigma$, $a \ne b$. Since $|w| = 2p + 2 \ge p$, there is some splitting $w = xyz$ satisfying the lemma. Since we know $|xy| \le p$, it must be that $x = a^k$ and $y = a^\ell$, with $k + \ell \le p$ and $\ell > 0$. This means that $z = a^{p-k-\ell} bb a^p$. Let $i = 0$, which gives $xz = a^{p - \ell} bb a^p$, which is not in $L$. By contradiction, $L$ is irregular.
By "splitting", I mean strings $x$, $y$, $z$ in $\Sigma^\ast$ such that $w = xyz$, $|xy| \le p$, and $|y| > 0$.
Best Answer
You’ve misunderstood closure under concatenation. The class of regular languages is closed under concatenation; individual regular languages in general aren’t. Thus, if the language $L$ is regular, so is the language $L\circ L$, but that does not mean that $0000\in L$ because $00\in L$.