$\binom{n}{1},\binom{n}{2},\ldots ,\binom{n}{n-1}$ are all even numbers. 
Consider the binomial coefficients $\binom{n}{k}=\frac{n!}{k!(n-k)!}\ (k=1,2,\ldots n-1)$. Determine all positive integers $n$ for which $\binom{n}{1},\binom{n}{2},\ldots ,\binom{n}{n-1}$ are all even numbers.

The answer is $n\in\{2^k|k\in\mathbb{N}\}$, but we have to show two things. First, if $n = 2^k$ then $\binom{n}{1},\binom{n}{2},\ldots ,\binom{n}{n-1}$ are all even numbers and second if $n \neq 2^k$ then $\binom{n}{1},\binom{n}{2},\ldots ,\binom{n}{n-1}$ are not all even numbers. I think we can use this for the first part but what about the second part?

As someone mentioned in the comments we'll use Lucas' Theorem and we'll show that if $n=2^km$ for some odd integer $m>1$, then we'll prove that $\binom{2^km}{2^k}$ is odd. We have:
$$2^km = a_t2^t + a_{t-1}2^{t-1} + ... + 1\cdot2^{k} + 0 \cdot 2^{k-1} +... + 0\cdot2^0$$
$$2^k = 1\cdot2^{k} + 0 \cdot 2^{k-1} +... + 0\cdot2^0$$
Now we have by Lucas' Theorem we have:
$$\binom{2^km}{2^k} \equiv \prod_{n=0}^{t} \binom{a_n}{b_n} \equiv 1 \pmod 2$$
This is true as all factors are equal to $1$, as they are either $\binom{0}{0}$, $\binom{1}{1}$ or $\binom{1}{0}$. Hence $\binom{2^km}{2^k}$ is odd.
Now if $n=2^k$, then we have one of the binomial coefficient zero, as $b_s= 1$ and $a_s=0$, for some $0 \le s \le k-1$. And as by definition $\binom{0}{1} = 0$ we have that: 
$$\binom{2^k}{q} \equiv \prod_{n=0}^{t} \binom{a_n}{b_n} \equiv 0 \pmod 2$$
for all $1 \le q \le 2^k - 1$. Hence all of them are even.
