For a fixed d, show that $n {{n-1} \choose d} p^d (1-p)^{n-1-d} \leq o(1)$. 
Given $p \in [0,1], pn^{(d+1)/d}  \to \infty$ and $np - \text{log}n - d \text{loglog}n \to \infty$, for a fixed d, show that $$n {{n-1} \choose d} p^d (1-p)^{n-1-d} \leq o(1), \text{as } n \to \infty.$$

I’ve been stuck with this for a day. Taking log of the first expression for $p$ would give log$n$, but I can’t see where that loglog$n$ comes from.
 A: From $\binom{n}{k}=\frac{n}{k}\cdot \binom{n-1}{k-1}$, we have
$$\color{blue}{n\cdot \binom{n-1}{d}\cdot p^d (1-p)^{n-1-d}}=\\
(d+1)\cdot\frac{n}{d+1}\cdot \binom{n-1}{d}\cdot p^d (1-p)^{n-1-d}=\\
\color{red}{\frac{d+1}{p}}\cdot\color{orange}{\binom{n}{d+1}\cdot p^{d+1}(1-p)^{n-(d+1)}}\tag{1}$$
Of course $p\ne0$, because $pn^{(d+1)/d}  \to \infty$ (otherwise it would $\to 0$).
Using binomial theorem
$$\color{orange}{\binom{n}{d+1}\cdot p^{d+1}(1-p)^{n-(d+1)}}\leq\\ \sum\limits_{k=0}^n\binom{n}{k}\cdot p^k(1-p)^{n-k}=\\
(p+1-p)^n=1\tag{2}$$
Adding $(1)$ and $(2)$ altogether
$$\color{blue}{n {{n-1} \choose d} p^d (1-p)^{n-1-d}} \leq \color{red}{\frac{d+1}{p}}\tag{3}$$
The RHS is a constant, so
$$n {{n-1} \choose d} p^d (1-p)^{n-1-d}=\mathcal{O}(1)$$

Update. However, the question is about $\mathcal{0}(1)$. Let's note by $b=\max\{p,1-p\}\in(0,1)$. From $(1)$
$$\color{blue}{n\cdot \binom{n-1}{d}\cdot p^d (1-p)^{n-1-d}}\leq\\
\frac{d+1}{p}\cdot\binom{n}{d+1}\cdot b^{d+1}b^{n-(d+1)}=
\frac{d+1}{p}\cdot\binom{n}{d+1}\cdot b^{n}\leq \\
\frac{d+1}{p}\cdot n^{d+1}\cdot b^{n}=
\frac{d+1}{p}\cdot \frac{n^{d+1}}{\left(\frac{1}{b}\right)^{n}}=...$$
noting $a=\frac{1}{b}>1$
$$...=\frac{d+1}{p}\cdot \frac{n^{d+1}}{a^{n}}\to0, n\to\infty$$
This post explains why.
A: Since $d$ is fixed, $0<p<1$ we can estimate
$$
n {{n-1} \choose d} p^d (1-p)^{n-1-d} \leq c\,n^{d+1}(1-p)^{n},
$$
which tends to zero, because $(1-p)^n$ exponentially small.
If I understand your problem correctly.
