I'm looking for a proof of the following hypothesis:
Given the tuple $(\{a_0, ..., a_{n-1}\},++)$, where $a_i, i = 1 ... n-1 \in \mathbb{N}$ represent arbitrary, pairwise different symbols and $++$ is an increment operator, such that
$$ \begin{array}{l} ++(a_{i_{k-1}}a_{i_{k-2}}...a_{i_1}a_{i_0}) = \begin{cases} ++(a_{i_{k-1}}a_{i_{k-2}}...a_{i_1})a_0 \quad \text{if } i_0 = n-1\\ a_{i_{k-1}}a_{i_{k-2}}...a_{i_1}a_{i_0+1} \quad \text{else}, \end{cases} \\ ++(a_i) = a_{i+1}, \quad i < n-1 \end{array} $$
where $a_ia_j$ is the concatenation of two symbols. Two objects $a_{i_{k-1}}a_{i_{k-2}}...a_{i_0}$ and $a_{j_{k-1}}a_{j_{k-2}}...a_{j_0}$, made up from concatenated symbols, are considered different, if $a_{i_l} \neq a_{j_l}$ for at least one $l \in \{0,...,k-1\}$.
Show that, after $p$ times repeated incrementation, starting from $\underbrace{a_0a_0...a_0}_m$, the concatenated object
$$(++)^p(a_0a_0...a_0), \quad 1 \le p < n^m$$
differs from every of its predecessors.
Proof by induction. Applying the $++$-operator $n^k$-times increments index $i_k$. This holds for $k=0$ from the definition of $++$.
($k \rightarrow k+1$): From the hypothesis, we have that incrementing $i_k$ $n$-times requires $\underbrace{n^k + ... + n^k}_n = n \cdot n^k = n^{k+1}$ incrementations. Where the definition of the incrementation operator for $i_k = i_{k-1} = ... = i_0 = n-1$ requires that
$$++(a_{i_{m-1}}a_{i_{m-2}}...a_{i_{k+1}})a_{n-1}...a_{n-1} = a_{i_{m-1}}a_{i_{m-2}}...a_{i_{k+1}+1}a_0...a_0,$$
which represents the $n^{(k+1)\text{st}}$ incrementation operation.
Since $a_{i_k+1} \neq a_{i_k} $ for $i_k < n-1$ from the definition of $a_i$, the concatenated object $a_{i_{m-1}}a_{i_{m-2}}...a_{i_k+1}...a_{i_0}$ differs from every of its predecessors.
Remark: The $n^m$-th incrementation operation cannot be applied due to overflow resulting in a total of $(n^m-1)+1 = n^m$ unique concatenated objects.