Under what minimal conditions on $d,c>0$, do we have $\lim_{n \to \infty}\sum_{n/2 \le k \le n}{n\choose k}(1-1/n^c)^{k\choose d} = 0$? Let $n$ and $d$ be positive integers and $c > 0$.

Question. Under what minimal conditions on $d$ and $c$ do we have $\lim_{n \to \infty}\sum_{n/2 \le k \le n}{n\choose k}(1-1/n^c)^{k\choose d} = 0$ ?

Intuitively, this should be the case if $c$ is sufficiently small compared to $d$. Just how small, that's the question...
 A: This holds if and only if $c<d-1$.
I’ll write $f(n)\sim g(n)$ for $g(n)=f(n)(1+o(1))$, that is, $\lim_{n\to\infty}g(n)/f(n)=1$. Also, $n/2$ stands for $\lceil n/2\rceil$. Let
$$S_{d,c}(n)=\sum_{k=n/2}^n\binom nk(1-n^{-c})^{\binom kd}$$
be the quantity we want to estimate. We have
$$\binom{n/2}d\sim\frac{n^d}{2^dd!}$$
and
$$\log(1-n^{-c})\sim-n^{-c},$$
thus
$$\log\left((1-n^{-c})^{\binom{n/2}d}\right)\sim-\frac{n^{d-c}}{2^dd!}.$$
Also, for $n/2\ge d$,
$$\binom n{n/2}(1-n^{-c})^{\binom{n/2}d}\le S_{d,c}(n)\le\sum_{k=n/2}^n\binom nk(1-n^{-c})^{\binom{n/2}d}\le2^n(1-n^{-c})^{\binom{n/2}d},$$
while
$$\log\binom n{n/2}\sim\log2^n=n\log2,$$
hence
$$\log S_{d,c}(n)=n(\log2+o(1))-\frac{n^{d-c}}{2^dd!}(1+o(1)).$$
Thus, if $d-c>1$, the $n^{d-c}$ term dominates, and
$$S_{d,c}(n)=\exp\left(-\frac{n^{d-c}}{2^dd!}(1+o(1))\right)\to0\qquad(n\to\infty),$$
whereas if $d-c<1$, the $n$ term dominates, and
$$S_{d,c}(n)=\exp\bigl(n(\log2+o(1))\bigr)\to+\infty\qquad(n\to\infty).$$
If $d-c=1$, we have
$$\log S_{d,c}(n)=\left(\log2-\frac1{2^dd!}+o(1)\right)n,$$
and since $\log2>1/(2^dd!)$ for $d\ge1$, again
$$S_{d,c}(n)\to+\infty\qquad(n\to\infty).$$
