Number of words in $L$ of length $n$?

Question

Be $L$ a formal language over the alphabet $\Sigma$. $L$ can be defined by induction:
$L^0 = \{\epsilon\}$ and for $i>0: L^i = \Sigma*L^{i-1}$.
Means that $|L^n|$ is the number of words in $L$ of length $n \in \mathbb{N}$ letters from $\Sigma$.
Find $|L^n|$.

For $\Sigma = \{a,b\}$, $|L^n| = 2^n$ scince $|\Sigma| = 2$ and $|K*\Sigma| = |K|*2$ (with $K$ being an arbitrary set). Right?
But what happens if $\Sigma = \{a,b,aa,ab,ba,bb\}$ ?
For some $n$ some elements can be created in multiple ways (e.g. $a$ and $ba$ result in $aba$ and $ab$ and $a$ result in $aba$). Therefore $|L^n| \neq 6^n$.
The problem clearly is, that some letters of $\Sigma$ (e.g. $aa$) consist of other letters of $\Sigma$ (e.g. $a$).
So far i looked at some results (no guaranty for correctness):

$|L^1| = 6$
$|L^2| = 28$
$|L^3| = 120$
$|L^4| = 496$
$|L^5| = 2016$
$|L^6| = 8128$

And found a formula for these values: $|L^n| = 2^n*(2^{n+1}-1)$
However, i can't explain or prove it. What is the correct way to approach this? Which words can be created in multiple ways?

Answer 1

Means that $|L^n|$ is the number of words in $L$ of length $n\in \mathbb{N}$ letters from $\Sigma$.

Actually, $L^n$ consists of the words that are composed of $n$ copies of words from $L$. The precise lengths of these words differ.

This exercise starts however with a very specific set $L$, it consists of all words of length 1 or 2. That means that $L^2$ consists of all words of length 2,3 or 4. (No need to verify in how many ways that can be done.) And this are 4+8+16 = 28 strings, exactly like you claimed.

This can be generalized for $L^n$.