Show using Stirling's approx. that $\log\binom{n}{\gamma n} = nh(\gamma) -\frac{1}{2} \log n + O(1).$

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Let $h(p) = -p \log p-(1-p)\log (1-p)$ denote the binary entropy of a Bernoulli distribution when the probability of observing a zero is $p$, where $\log$ denotes the logarithm to base 2. Show, using Stirling's approximation:

$$ \ln(n!) = (n+\frac{1}{2})\ln(n) - n + O(1)$$

that the following holds:

$$\log\binom{n}{\gamma n} = nh(\gamma) -\frac{1}{2} \log n + O(1).$$

How should I do this? I'd like a hint to get me started.