Is there a combinatorial explanation for this identity related to Kraft's inequality? Kraft's inequality involves the quantity:
$$\sum_{x \in X} \frac 1 {b^{\ell(x)}} \tag 1$$
Where we are considering a code mapping symbols in the alphabet $X$ to strings in an alphabet of $b$ symbols. $\ell(x)$ denotes the number of symbols in the endoding of the symbol $x\in X$. So if $b=2$ and $x$ is encoded as $11001$, then $\ell(x)=5$.
In a common proof of Kraft's inequality, we raise this quantity to the $n$-th power, and it turns out we have the following identity:

$$\left(\sum_{x \in X} \frac 1 {b^{\ell(x)}}\right)^n=\sum_{x \in X^n} \frac 1 {b^{\ell(x)}}$$

Where $X^n$ denotes the set of all strings over $X$ of length exactly $n$. This is a remarkable identity that can be proven algebraically by expanding the left hand side, gathering terms, and applying basic properties of exponents.
My question is: is there a combinatorial explanation for this? That is, can the quantity $(1)$ and the operation of raising it to the $n$-th power be given combinatorial interpretations so that the identity is more obvious?
 A: Not sure whether this addresses your question, but write
$$
\sum_{x\in X}\frac{1}{b^{\ell(x)}}=\sum_{x\in X}\left(\frac{1}{b}\right)^{\ell(x)}=\sum_{x\in X}s^{\ell(x)},
$$
where $s=1/b.$  For the moment we will ignore that $s$ equals $1/b$ and just treat $s$ as a symbol.  Then if one collects terms in the sum, the coefficient of $s^j$ will be the number of elements of $X$ whose representation has length $j.$  In other words, the sum is the length generating function for $X.$
Now the set $X^2$ is formed by concatenating two strings from $X$ in all possible ways.  The length generating function for $X^2$ will simply be the square of the length generating function for $X$.  I hope it is clear why this is true.  Suppose, for example that $X=\{110,111,1100\}.$  Then
$$
X^2=\{110|110,110|111,110|1100,111|110,111|111,111|1100,1100|110,1100|111,1100|1100\}.
$$
where the vertical bar has been used to indicate the point of concatenation.  The length generating function for $X$ in this example is
$$
2s^3+s^4=sss+sss+ssss.
$$
If we multiply this by itself, we get
$$
\begin{aligned}
(sss+sss+ssss)(sss+sss+ssss)&=sss*sss+sss*sss+sss*ssss\\
&\,+sss*sss+sss*sss+sss*ssss\\
&\,+ssss*sss+ssss*sss+ssss*ssss\\
&=4s^6+4s^7+s^8.
\end{aligned}
$$
The process of expanding the product above using the distributive law completely parallels the process by which $X^2$ was formed.  Imagine that in the $X^2$ the symbols "$0$" and "$1$" are replaced with "$s$", the symbol "$|$" is replaced with "$*$", and the symbol "$,$" is replaced with "$+$".  Then one gets the length generating function for $X^2.$
What is true of $X^2$ is true of $X^n,$ for any positive $n.$
Added: The coefficient of $s^j$ in $\sum_{x\in X^n}s^{\ell(x)}$ is the number of strings of length $j$ in $X^n.$  The coefficient of $s^j$ in $\left(\sum_{x\in X}s^{\ell(x)}\right)^n$ is the number of $n$-fold sums that equal $j$ with terms taken from the multiset of lengths of strings in $X.$  Since $X^n$ consists of all $n$-fold concatenations of strings in $X,$ these coefficients must be equal.
If you have $k$ things, then $k^n$ is the number of ordered length-$n$ strings of those things, with repetition allowed.  The only new ingredient here is that $k$ is a generating function rather than a number.  This allows us to refine the counting—in this case, by length.  So if we have $k=c_0+c_1s+\ldots+c_rs^r,$ then we have $c_0+c_1+\ldots+c_r$ things, but $c_0$ of them are of length $0,$ $c_1$ are of length $1,$ and so on.  Now when we take the $n^\text{th}$ power of $k,$ we have $(c_0+c_1+\ldots+c_r)^n$ ordered length-$n$ strings of those things, with repetition allowed, but, in addition, if we look at the coefficients of powers of $s,$ we have the numbers broken down by length.
For certain special values of $s,$ the equality is more straightforward to interpret.  If one sets $s=1,$ then $\sum_{x\in X}s^{\ell(x)}=\lvert X\rvert$ is the cardinality of $X.$  The $n^\text{th}$ power of this quantity, $\lvert X\rvert^n,$ is the number of $n$-tuples with elements taken from $X.$  But, by definition of $X^n,$ so is the sum $\sum_{x\in X^n}s^{\ell(x)}=\lvert X^n\rvert.$
We can interpret $s=2$ if, when each element of $X$ is encoded as a string in the new alphabet, we allow each letter to be colored in one of two colors.  Then $\sum_{x\in X}s^{\ell(x)}$ is the cardinality of the set of colored strings of $X,$ and $\sum_{x\in X^n}s^{\ell(x)}$ is the cardinality of the set of colored of strings of $X^n.$  But by definition, the latter is obtained by stringing together colored strings from $X,$ which establishes the equality.
The same thing works for any positive integer $s$ by increasing the number of colors.  It’s hard to see what the combinatorial interpretation might be when $s$ takes a non-positive-integer value, such as $1/b,$ since the sums will no longer be counting anything.  But I don’t see why you need to find such an interpretation.  The combinatorics is all there in the general case where $s$ is an indeterminate.
