The Lehmer code is a bijection $S_n\to {\prod }_{i=1}^n\{0,\dots ,i-1\}$

For $\sigma \in S_n$, the component ${\mathrm{\ell }}_i(\sigma )$ counts entries to the right of position $n-i$, and there are exactly $i-1$ such positions, so $0\le {\mathrm{\ell }}_i(\sigma )\le i-1$. Thus $L(\sigma )$ lies in the stated codomain.

Conversely, let $(c_1,\dots ,c_n)$ lie in the codomain. Begin with the increasing list $R=[0,1,\dots ,n-1]$ of unused values. For $i=n,n-1,\dots ,1$, put in position $n-i$ the value of $R$ having exactly $c_i$ smaller elements in $R$, and delete that value from $R$. This choice exists and is unique because $R$ has $i$ elements and $0\le c_i<i$. Call the resulting permutation ${\sigma }_c$.

The permutation ${\sigma }_c$ has Lehmer code $(c_1,\dots ,c_n)$. At the stage for $i$, all values then left in $R$ will occupy the $i-1$ positions to the right of position $n-i$. The chosen value has exactly $c_i$ smaller values in $R$, so exactly $c_i$ later entries of ${\sigma }_c$ are smaller than it. This is the defining equality ${\mathrm{\ell }}_i({\sigma }_c)=c_i$.

The construction is unique. At position $0$, all $n$ values are unused and the code coordinate $c_n$ requires the entry with exactly $c_n$ smaller unused values. After deleting it, coordinate $c_{n-1}$ uniquely determines the next entry among the remaining values, and so on. Thus any permutation with code $c$ is exactly ${\sigma }_c$. Step 2.1 gives surjectivity and this uniqueness gives injectivity.

Therefore $L$ is a bijection. The case $n=0$ is included: both sides are the singleton empty object.