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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?

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    The induction idea is good (it's not how people usually prove this, but it's perfectly valid). Now, what is $\dbinom{p}{k+1} / \dbinom{p}{k}$ ? When does dividing by some integer preserve divisibility by $p$ ? – 2017-01-23
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    Minor gripe: You don't want $a \in \mathbb{R}$. You want $a$ to be an independent variable (as in, the generator of the polynomial ring $\mathbb{Z}\left[a\right]$). It is not enough to have $\left(1+a\right)^p = 1 + c_1 a + c_2 a^2 + \cdots + a^p$ to hold for **some** choice of real $a$. – 2017-01-23
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    @darijgrinberg thanks for your comments. Per your first question, $\frac{p-k}{k+1}$. If it is indeed true that $p$ divides this, then I can finish the proof. However, I'm not sure I understand what you mean when you say "When does dividing by some integer preserve divisibility by $p$? Thanks for your help so far. – 2017-01-23
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    Well, $p$ does not divide $\dfrac{p-k}{k+1}$; after all, it's not even an integer. But let's see it as follows: You know that $\dbinom{p}{k+1}$ is obtained from $\dbinom{p}{k}$ by first multiplying by $p-k$ and then dividing by $k+1$. The first step clearly preserves divisibility by $p$. Does the second, too? – 2017-01-23

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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.