Show that two different representations to the base $k$ represent two different integers I would like to show:

Given two distinct, positive, integer representations in base $k$, say $\sum_{i=0}^na_ik^i$ and $\sum_{i=0}^mb_ik^i$ where $a_n \neq 0 \neq b_m$ and $a_i,b_i \in \{0,1,\ldots , k-1 \}$, prove that $$\sum_{i=0}^na_ik^i \neq \sum_{i=0}^mb_ik^i$$

I would also like to show this using the result that $$\sum_{i=0}^pc_ik^i \leq k^{p+1}-1$$ for every integer representation in base $k$. Additionally, I don't want to use the Basis Representation Theorem (that every basis representation is unique). 
What I have so far: I figured there are two cases to make $\sum_{i=0}^na_ik^i$ and $\sum_{i=0}^mb_ik^i$ be distinct. First, if WLOG $m>n$. Then we know $$\sum_{i=0}^na_ik^i\leq k^{n+1}-1 \leq k^m-1 < k^m \leq \sum_{i=0}^mb_ik^i$$ Then I moved on to the second case of $m=n$. For the two integer representations to be distinct then there must be some $i \in \{1,2,\ldots , n \}$ such that $a_i \neq b_i$. At this point I am stuck on how to show the two integer representations must be different numbers, using the result of $\sum_{i=0}^pc_ik^i \leq k^{p+1}-1$. Does anyone have an idea how to do this? Or to do away with cases?

So we have $A = \sum_{0 \leq i \leq n} a_ik^i = \sum_{0 \leq i \leq m} b_ik^i = B$. Let's assume WLOG $n \geq m$. If $n = m$ then subtract $b_mk^m$ from both sides to make the RHS have lesser degree. Once the RHS has lesser degree, subtract $\sum_{0 \leq i \leq n-1} a_ik^i$ from both sides to get something of the form
$$ak^n = \sum_{0 \leq i \leq n-1} c_ik^i$$
where $a \in \{0, ... , k-1\}$ and $c_i \in \{-(k-1), ... , k-1\}$.
We will now show that $a$ must be $0$, otherwise our LHS is strictly greater than our RHS.
Assume $a > 0$. Then we can see the maximal value of the RHS, $\sum_{0 \leq i \leq n-1} c_ik^i$, is when the $c_i$ are all $k-1$ (or $-(k-1)$ possibly for odd degree terms but this doesn't change the maximal value) giving a maximum of
\begin{align*}
(k-1) + (k-1)k + (k-1)k^2 + ... + (k-1)k^{n-1} &= (k-1)(1 + k + k^2 + ... + k^{n-1}) \\
&= (k-1)\frac{k^n - 1}{k-1} \\
&= k^n - 1
\end{align*}
which is of course strictly smaller than the minimum for our LHS, $k^n$. Hence $a = 0$ which means either $a_n = 0$ (if $A$ had a greater degree than $B$) or $a_n = b_n$ (if they had equal degree).
