The problem goes like this: Let $N\leq e^{c/ \epsilon}$ and $0\leq j\leq \log_2 N^2$. Suppose for some $C>0$, we have $$\mathbb{E}(\max_{A\in\mathit{A_j}}h_A)<C j \epsilon 2^j,$$ where $A\subset \mathbb{Z}^2$, $A_j$ just a collection of certain $A$, and $h_A=\sum_{v\in A}h_v$ with each $h_v$ i.i.d Gaussian random variables with mean $0$ and variance $\epsilon^2$. If this is true, then by Gaussian concentration inequality, we have for $c>0$ small enough, $$\mathbb{P}(\max_{A\in \mathit{A_j}}h_A\geq 2^j)\leq e^{-c/{\epsilon^2}}.$$
So my question is how to get the second inequality by using Gaussian concentration inequality, which I couldn’t find online.
My attempt is the following: $$\begin{aligned}\mathbb{P}(\max_{A\in\mathit{A_j}}h_A\geq 2^j)&=\mathbb{P}(e^{\max_{A\in\mathit{A_j}}h_A}\geq e^{2^j})\\ &\leq \frac{\mathbb{E}(e^{\max h_A})}{e^{2^j}} \end{aligned}$$ But to get $\mathbb{E}(e^{\max h_A})$, we need higher moments of $\max h_A$. So is this the point where Gaussian concentration inequality comes?
Thank you very much!