$p$ is prime and $n>1$ integer such that $p|{n \choose k} , \forall 1\le k \le n-1$ , then $n$ is a power of $p$? Let $p$ be a prime and $n&gt;1$ be an integer such that $p|{n \choose k} , \forall 1\le k \le n-1$ , then is it true that $n$ is a power of $p$ ?

From the given condition, I am only able to derive that $p|n$ . (I know that the converse of the claim I make , is true)

Please help.
HINT:

Indeed, see Lucas theorem

ADDED:

Lucas' theorem states: If $n= \sum a_i p^i$ and $k=\sum b_i p^i$ are base $p$ expansions then 
$$\binom{n}{k} \equiv \prod \binom{a_i}{b_i} \pmod p$$

Now if $n$ is not a $p$ power there exists $1\le k &lt;n$ so that $b_i \le a_i$ for all $i$. Note that $a_i \le p-1$. We get $\binom{n}{k} \not \equiv 0 \pmod p$.
HINT:

Indeed, see Lucas theorem

ADDED:

Lucas' theorem states: If $n= \sum a_i p^i$ and $k=\sum b_i p^i$ are base $p$ expansions then 
$$\binom{n}{k} \equiv \prod \binom{a_i}{b_i} \pmod p$$

Now if $n$ is not a $p$ power there exists $1\le k &lt;n$ so that $b_i \le a_i$ for all $i$. Note that $a_i \le p-1$. We get $\binom{n}{k} \not \equiv 0 \pmod p$.