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?
 A: Without loss of generality, assume $n \geq m$. Let $A = \sum_0^na_ik^i$ and 
$B = \sum_0^mb_ik^i$.
If $\forall i \leq m: a_i = b_i$ then $A-B = \sum_{i=m+1}^n a_ik^i \geq a_nk^n > 0$ since $a_n > 0$. Thus we need only consider the case where $\exists i \leq m: a_i \neq b_i$.
Let $j = \inf i : a_i \neq b_i$ and let $D = \sum_{i=0}^{j-1} a_ik^i = \sum_{i=0}^{j-1} b_ik^i$.  Note that
$$ 0 < |a_j - b_j| < k$$
Let 
$$
A' = A-D =  a_jk^j + \sum_{i=j+1}^{n}$ a_ik^i \\
B' = A-D =  \sum_{i=j+1}^{m}$ a_ik^i 
$$
(Here, the sum for $B'$ is as usual considered to be zero if $j+1>m$.)
$A \neq B$ if and only if $A' \neq B'$.  Since each $k^r$ for $r \geq j+1$ is divisible by $k^{j+1}$,
$$
\sum_{i=j+1}^{n} a_ik^i = p_a k^{j+1} \\
\sum_{i=j+1}^{n} b_ik^i = p_b k^{j+1}
$$ 
for $p_a, p_b \in \Bbb{N}$.  Thus
$$
A' - B' = (p_a-p_b) k^{j+1} + (a_k-b_k)k^j = k^j \left[ (p_a-p_b) k + (a_j-b_j) \right] 
$$
So $\frac{A' - B'}{k^j}$ is an integer, and it is a  non-zero integer multiple of $k$ plus a non-zero integer of absolute value less than $k$.  Thus
$$\frac{A' - B'}{k^j} \neq 0 \Longrightarrow A'-B' \neq 0 \Longrightarrow A \neq B.$$
QED
A: I will do the case for $n=m$ since graydad has done the work for n>m WLOG. 
Let $A=\sum_{i=0}^na_ik^i$ and $B=\sum_{i=0}^mb_ik^i$.
 If A=B, then We should have $A-B=B-A=0$. 
$A-B=$[$a_nk^n+a_{n-1}k^{n-1}+...+a_0$]$-[b_nk^n+b_{n-1}+...+b_0]$$=$$(a_n-b_n)k^n+(a_{n-1}-b_{n-1})k^{n-1}+...+(a_0-b_0)=\sum_{i=0}^n(a_i-b_i)k^i$. 
If $∀i$  $(a_i-b_i)<0$, then $\sum_{i=0}^n(a_i-b_i)k^i<0$  and we have $A \neq  B$. Same reasoning for $(a_i-b_i)>0$, $∀i$. If $∀i$  $(a_i-b_i)=0$, then the representations are identical.    
Therefore, we must have some terms w/ positive and some w/ negative coefficients.
We can group all the terms with negative coefficients together and all the positive ones together. 
Let $j$ be the largest exponent for the terms w/ positive coefficients and $t$ the largest for the negative coefficients. Therefore, we have $A-B=$[$a_nk^n+a_{n-1}k^{n-1}+...+a_0$]$-[b_nk^n+b_{n-1}+...+b_0]$$=$$(a_n-b_n)k^n+(a_{n-1}-b_{n-1})k^{n-1}+...+(a_0-b_0)=\sum_{i=0}^n(a_i-b_i)k^i$=$(a_j-b_j)k^j+$ (rest_of_terms w/ positive coefficients)$-[- (a_t-b_t)k^t+$(rest of terms w/ negative coefficients)]
Note that $A-B$ is now written as the difference of two integers in base k representations of differing degrees, $j\neq t$. 
So let $A'=(a_j-b_j)k^j+$ (rest_of_terms w/ positive coefficients) and $B'=[- (a_t-b_t)k^t+$(rest of terms w/ negative coefficients)]. This means, $A-B=A'-B'$
WLOG let $t>$j, Then $A'\leq k^{j+1}-1< k^{t} \leq B'$ $\Longrightarrow A\neq B$.
(done)
A: This is a comment,
not an answer,
but the comment space is too small
to hold this.
I proved this result on
representation in general bases
over 40 years ago:
Let $\mathbb{B} =(B_j)_{j=0}^{\infty}$
be an increasing series
of positive integers
with $B_0 = 1$.
A positive integer $n$
is represented in
$\mathbb{B}$ if
$n$ can be written in the form
$n 
= \sum_{j=0}^{\infty} d_j B_j
$
where
$0 \le d_j < B_{j+1}/B_j
$.
(For the usual representation,
let $B_j = b^j$
for some integer $b \ge 2$.)
Then
(1) Every positive integer
can be represented in a form with
all but a finite number of the
$d_j$ being zero.
(2) The representation is unique
for all positive integers
if and only if
$B_j$ divides $B_{j+1}$
for all $j$.
Uniqueness then holds
in the case mentioned above
($B_j = b^j$),
but also holds in the factorial numbering system,
where
$B_j = (j+1)!$.
A: 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
A: 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).
