Let L be any non-empty language over an alphabet Σ. Show that $L^2$ ⊆ $L^3$ if and only if λ ∈ L I have this question in my Theoretical Computer Science class on the topic of Automata and Formal Languages.

$$ λ ∈ L \iff L^2 ⊆ L^3$$

I thought of showing it by proving the contrapositive but I was only able to complete half of the proof (since it is a bidirectional proof) and on top of that I am not even sure it is correctly written in a formal form. 

I wanted to prove that if 

$$λ ∉ L \implies L^2 \nsubseteq L^3$$

But since it is a bidirectional proof, I must also show that 

$$L^2 \nsubseteq L^3 \implies λ ∉ L$$

right?

This is what I have so far:

1) $λ ∉ L \implies L^2 \nsubseteq L^3$

$ \lambda \notin L \implies \forall w \in L, n_o(w) \geq 1$
$ L^2 = L * L \implies \forall w_2 \in L^2, n_o(w_2) \geq 2 $
$ L^3 = L * L * L \implies \forall w_3 \in L^3, n_o(w_3) \geq 3 $
$\forall w_3 \in L^3 \land \forall w_2 \in L^2, n_o(w_3) &gt; n_o(w_2)$
$ \therefore L^2 \nsubseteq L^3$

So first of all, is this correct? 

Secondly, how would I go about showing the second part of my proof where
$$L^2 \nsubseteq L^3 \implies λ ∉ L$$

Thank you!
I'm assuming by $\lambda$ you mean the empty word and by $n_0(w)$ the length of a word.

The proof for the first part is not correct: You argue that every word in $L^2$ has length at least $2$ and every word in $L^3$ has length at least $3$. From that it does not follow that every word in $L^3$ is longer than every word $L^2$, because there could also be a word of e.g. length $3$ in $L^2$. Consider for example $L = \{a\}^\star \setminus \{ \lambda \}$, that is, the set of all non-empty $a$-sequences. You have $a \in L$ and $aa \in L$, thus $aaa = a \cdot aa \in L^2$, but also $aaa = a \cdot a \cdot a \in L^3$.

Instead, what you want to say is that there is a word of minimal length $n$ in $L^2$, and because every word in $L$ has length at least one, the shortest word in $L^3$ has length at least $n+1$.

More formally: Let $\lambda \notin L$. Thus $\forall w \in L: n_0(w) &gt; 0$. Pick $w \in L^2$ such that $n_0(w)$ is minimal. Assume $w \in L^3$. Since $L^3 = L^2 \cdot L$, it follows that $w = u \cdot v$ for $u \in L^2$ and $v \in L$. Since $n_0(v) &gt; 0$, it follows $n_0(w) &gt; n_0(u)$. But then $n_0(w)$ is not minimal. Contradiction.

For the second part: Assume $L^2 \not \subseteq L^3$ and $\lambda \in L$. Thus there is $w \in L^2$ such that $w \notin L^3$. But since $L^3 = L^2 \cdot L$, it follows $w = w \cdot \lambda \in L^3$. Contradiction.