Elementary number theory problem: If $m$ is a prime number and $a,b$ two numbers less than $m$, then prove the following Problem:

If $m$ is a prime number and $a,b$ two distinct positive integers less than $m$ then
  prove that$$a^{m-2}+a^{m-3}b+a^{m-4}b^2+\cdots +b^{m-2}$$is a multiple
  of $m$

Source:
Found it on a mock test paper. I have just started learning number theory (thanks fleablood) for an exam, and I'm learning all of it on my own. The problem looks like a simple one but I can't seem to solve it.
My try:
The given expression looks very similar to binomial expansion of $(a+b)^{m-2}$
$$(a+b)^{m-2} = {{m-2}\choose{0}}a^{m-2}+{{m-2}\choose{1}}a^{m-3}b+\cdots+{{m-2}\choose{m-2}}b^{m-2}$$or
$$(a+b)^{m-2} = a^{m-2}+{{m-2}\choose{1}}a^{m-3}b+\cdots+b^{m-2}$$
Now I have read a theorem which states that 
The coefficient of every term in the expansion of $(a+b)^{p}$, except the first and the last, is divisible by $p$ where $p$ is a prime number
Which means that all the middle terms (except first and last) of the binomial expansion are divisible by $m$. But what about the first and last terms? And also the function given in problem is not a binomial expansion. How do I proceed? All help appreciated!
 A: If $a\not=b$ then
$$a^{m-2}+a^{m-3}b+a^{m-4}b^2+\cdots +b^{m-2}=\frac{(a^{m-1}-1)-(b^{m-1}-1)}{a-b}.$$ 
which is divisible by the prime $m$ because, by the Fermat's little theorem, $m$  divide $(a^{m-1}-1)$ and $(b^{m-1}-1)$ while $0<|a-b|<m$. 
P.S. The property is false for $a=b$ because the prime $m$ does not divide
$$a^{m-2}+a^{m-3}b+a^{m-4}b^2+\cdots +b^{m-2}=(m-1)a^{m-2}.$$
A: I let you conclude why it is wrong for $a=b$ from the theory below.
Note that you expression is equivalent to $\frac{a^{m-1}-b^{m-1}}{a-b}$.
Since $a,b$ are smaller than $m$, we only need to prove that $a^{m-1} -b^{m-1}$ is divisible by $m$. Which is true (using Fermat little theorem) since $a,b$ are smaller  than $m$ and hence coprime to it as m is a prime .
A: Note that the the condition only holds for $a\neq b$    
We have $$a^{m-1}-b^{m-1}=(a-b)(a^{m-2}+a^{m-3}\ b+\ldots+b^{m-2}\ )$$    
We also have, by Fermat's Little Theorem, that $$a^{p-1}\ \equiv 1\pmod p,$$ where $p$ is a prime, and $p\nmid a$.  
It therefore follows that $$(a-b)(a^{m-2}+a^{m-3}\ b+\ldots+b^{m-2}\ )\ \equiv 0\pmod m$$$$\Rightarrow m\mid (a-b)\;\; OR\;\;m\mid (a^{m-2}+a^{m-3}\ b+\ldots+b^{m-2}\ )$$
The aboce follows from that $m$ is a prime.  
Since we have $a,b\lt n$, and $a\neq b$, we conclude $m\nmid (a-b)$
$$\Rightarrow m\mid (a^{m-2}+a^{m-3}\ b+\ldots+b^{m-2}\ )$$
