Combinatorial inequality $\binom{n}{j}\leqslant 2^n$ I was trying to prove (or to find a counterexample) of the following inequality:
$$\binom{n}{j}\leqslant 2^n$$
As I coudn't find a proof/counterexample, I tested some numbers and could see it working up to $n=12$. Does anyone have a proof/counterexample of it? I've tried to pass the $\log_2$ in both sides, in order to use properties of $\log$, but I could't find anything better. Any help will be appreciated.

Hint:


*

*$\displaystyle\sum_{j=0}^n \binom n j=2^n$

*$\displaystyle \binom n j, j \in [n]$ is the number of ways of choosing $j$ objects from $n$ of them. Can this be zero or negative?
