I am trying to evaluate the following limit: $$ \forall p\in(0,1) ,\lim_{n\rightarrow \infty} \frac{1}{n}\log{n\choose \lfloor np \rfloor} =H(p),$$ where $\lfloor x\rfloor$ means the greatest integer less than $x$ and $H(p)=-p\log(p)-q\log(q),q+p=1$ is the entropy of Bernoulli(p).
I know it could be evaluated by Stirling's formula, but the exercise asks not to use it.
One hint is using $$\forall \epsilon >0,\lim_{n\rightarrow \infty}P(|p^{\frac{S_n}{n}}q^{1-\frac{S_n}{n}} -e^{-H(p)}|>\epsilon)=0,$$ where $S_n=\sum_{i=1}^{n}X_i \;\text{and}\; X_i \;\text {iid, Bernoulli(p)}$. This follows from strong law of large number.
Since $S_n\sim B(n,p)$, we could have a term involving ${n\choose \lfloor np \rfloor}p^{\lfloor np\rfloor}q^{n-\lfloor np \rfloor}$ in computing $P(|p^{\frac{S_n}{n}}q^{1-\frac{S_n}{n}} -e^{-H(p)}|<\epsilon)$, but then how to proceed?
Any help would be appreciated!