I don't understand how we get the second line from the first line in the above image. Here $k = \delta \frac{d}{t}$ where $\delta$ is some very small constant greater than $0$. Also $N = d^3$.
So, last line is basically $2^{\Omega( \frac{d}{t}\log d)}$.
From the following expression $$ \frac{1}{2^{O(d/t)}} \operatorname{min}( {d \choose k}^2.d^k.k!.e^{(-kt)\ln\frac{1+ (k/d) \ln d}{1 - (k/d)\ln d}}, e^{(d-k-kt)\ln\frac{1+ (k/d) \ln d}{1 - (k/d) \ln d}}) $$
I can get
$$ \frac{1}{2^{O(d/t)}} e^{(-kt)\ln\frac{1+ (k/d) \ln d}{1 - (k/d)\ln d}} \operatorname{min}( {d \choose k}^2.d^k.k!., e^{(d-k)}) $$
But I don't understand how the above expression will give us $2^{\Omega(d/t \log d)}$