Doubt on $\ln k!$
Notice that $\ln n=\int_{n-1}^n\ln n\mathop{dx}$ for all $n=2,3,\dots, k$. Therefore,
\begin{align*} \ln 1+\ln 2+\ln 3+\dots+\ln k&=0+\int_{1}^2\ln 2\mathop{dx}+\int_{2}^3\ln 3\mathop{dx}+\dots+\int_{k-1}^k\ln k\mathop{dx}\\ &\ge \int_{1}^2\ln x\mathop{dx}+\int_{2}^3\ln x\mathop{dx}+\dots+\int_{k-1}^k\ln x\mathop{dx}\\ &=\int_{1}^k\ln x\mathop{dx}. \end{align*}
This inequality occurs since $\ln x$ is a strictly increasing function.