I am trying to understand the proof to a random graph problem (the threshold for connectivity of $G \sim G(n,p)$ being $\frac{logn}{n}$). I am struggling to see exactly why the following holds:
$\text{When } p^{*} \ll p,\\ n(1-p) ^{n-1} \leq ne^{-p(n-1)} = o(ne^{-\log n}) = o(1).\\$
I am aware that the following inequalities may be used in the derivation but my further deductions (after the $\implies$ symbol) show that the LHS of ($\text{*}$) and RHS of ($\text{**}$) are both greater or equal to the same quantity. However, the line of the proof provided above states something different, to which I am unable to arrive.
$\text{(*) }(1-p)^{n-1} \geq 1 - (n - 1)p, \forall 0 \leq p \leq 1, \text{and}\\ \text{(**) }e^{p} \geq p + 1, \forall p \implies e^{-p(n-1)} \geq 1-(n-1)p.$
Another amateurish question follows here - is the following the case in the last part of the aforementioned line from the proof: $ne^{-\log n} = nn^{-1} \text{?}$
$$1-p\leqslant\mathrm e^{-p}\implies(1-p)^n\leqslant\mathrm e^{-np}\implies n(1-p)^n\leqslant n\mathrm e^{-np}\quad\&\qquad\mathrm e^{-c\log x}=x^{-c}$$ Edit: The starting point $1-p\leqslant\mathrm e^{-p}$ (worth remembering in other contexts as well) can be proven in a wealth of different ways. For example, it follows from the fact that $\mathrm e^{-x}\leqslant1$ for every $x\geqslant0$ hence, for every $p\geqslant0$, $$ p=\int_0^p\mathrm dx\geqslant\int_0^p\mathrm e^{-x}\mathrm dx=1-\mathrm e^{-p}. $$