Induction proof: Show that $p\mid c_i \forall i$ such that $1\leq i \leq p-1$. 
Let $p$ be a prime, $a\in\mathbb{R}$ and suppose that we have:
  $$(1+a)^p=1+c_1a+c_2a^2+\cdots+c_{p-1}a^{p-1}+a^p$$
  Show that $p\mid c_i, \forall i$ such that $1\leq i \leq p-1$.

My work so far:
I think it is best to prove by induction.
First, note that by the binomial theorem,
$$(1+a)^p=1+\binom p1 a+\binom p2 a^2 + \cdots +\binom p{p-1}a^{p-1}+a^p$$
Therefore, $c_1=\binom p1=p$. Since $p\mid p$, the basic case is true.
In the inductive step I begin to run into problems. After assuming $p\mid\binom pk$, I can't seem to manipulate the algebra to show $p\mid\binom p{k+1}$.
Am I on the right track with this? Is induction the way to go about showing this property?
 A: It may help to write out $p \choose k$ in its expanded representation. That is to say: 
$$
{p \choose k} = \frac{p!}{k!(p - k)!} \\
{p \choose k + 1} = \frac{p!}{(k+1)!(p-k-1)!} = \frac{p!\cdot(p-k)}{(k+1)\cdot k!\cdot (p-k)!} = \frac{p!}{k!(p-k)!}\cdot\frac{p-k}{k+1}\\
= {p \choose k} \cdot \frac{p - k}{k + 1}
$$
and by the induction hypothesis we know that $p | {p \choose k}$. 
As a question, if $p | A$, and you take some other integer $B$ then what can you say about $p | AB$ (without knowing anything about $B$)? 
That should be sufficient to finish the proof. 

As an alternative means of proving this proposition, you could have tried to use a contradiction. So for instance:
Assume that $p \not{|} {p \choose k}$ for some $k \in [1, p-1]$. Then we must have that: $p \not{|} \frac{p!}{(p - k)! \cdot k!}$. 
$$
 \frac{p!}{(p - k)! \cdot k!} = \frac{p \times (p-1) \times ... \times (p - k)!}{(p - k)! k!} \\
= \frac{p \times (p - 1)\times ... \times (p - k +1)}{k!}
$$
Now, let's note that $p \choose k$ is an integer (this can be readily proven). 
Since we assumed that $p \not{|} \frac{p!}{(p - k)! \cdot k!}$, then we must have that $p | k!$ since $p$ clearly does divide into the numerator of the expression (it also must divide into the denominator for it not to be a factor). 
Expanding the product of $k!$ results in: 
$$k! = k \times (k - 1) \times (k - 2) \times ... \times 2 \times 1$$
So since $p | k!$, and $p$ is prime, we must have that one of these terms is divisible by (or equal to) $p$. But note that $p > k$ and so $p \not{|} x \in \mathbb{Z}_{\leq K}$. 
Thus we have reached a contradiction (namely that $p$ must divide a number which is smaller than it, or $p$ must not be prime), and as such our assumption that $p \not{|} {p \choose k}$ for some $k \in [1, p-1]$ must be false.
