Let $X_i \stackrel{iid}{\sim} \text{Binom}(n,p)$ for any $1 \leq i \leq n$ (assume that $p=o(\log n/n)$). Prove that $$ \Pr(\max_{1 \leq i \leq n} X_i \geq k^*) \rightarrow 1 \text{ as } n \rightarrow \infty, $$ where $k=\frac{\log n}{\log(\log n/(np))}$.
I tried the following approach but failed to get anything useful: $$ \Pr(\max_{1 \leq i \leq n} X_i < k^*) = (\Pr(X_1 < k^*))^n, $$ where \begin{align*} \Pr(X_1 < k^*)=\sum_{j=0}^{k^*-1} \binom{n}{j} \cdot p^j \cdot (1-p)^{n-j}<\sum_{j=0}^{k^*-1} n^j \cdot p^j \cdot (1-p)^{n-j}=(1-p)^{n} \sum_{j=0}^{k^*-1} \left(\frac{np}{1-p}\right)^j=... \end{align*}