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?

Case 2: We can assume $m = n$, $a_n \neq 0 \neq b_n$
IF $\;a_j \ne b_j$ for some $j$ $\;$ THEN
(1) $\;\sum_{i=0}^na_ik^i \neq \sum_{i=0}^nb_ik^i$
PROOF
If $a_n =  b_n$, then you can just drop off that 'highest degree' term. You can keep doing this and thereby just assume from the start that $a_n \ne b_n$. Suppose $a_n \gt b_n$, and let $c = a_n - b_n$ be the positive natural number representing the difference of those numbers. You can now reduce both sides of (1) by $b_nk^n$. So, you now must show that
$\sum_{i=0}^{n-1}a_ik^i + ck^n \neq \sum_{i=0}^{n-1}b_ik^i$
Now this is funny! We are back trying to show distinctness for Case 1, which has already been taken care of by the questioner. 
QED
