$\binom{n}{k}$ for $n>k^{100}$ This paper I'm reading says this
"We used the fact that $4k^2n^{k-1}<4(k+1)!\binom{n-1}{k-1}$ for $n>k^{100}$." The only condition here is that $k\geq3$ and $n$ can be sufficiently large. However, this seems incorrect to me.

Note that
$$\begin{align}\frac{(k-1)!{n-1\choose k-1}}{n^{k-1}}
&=\frac{(n-1)!}{n^{k-1}(n-k)!}\\
&>\frac{(n-k+1)^{k-1}}{n^{k-1}}\\
&=\left(1-\frac{k-1}{n}\right)^{k-1}\\
&>1-\frac{(k-1)^2}n \end{align}$$(where we first use that $\frac{(n-1)!}{(n-k)!}$ consists of $k-1$ factors $\ge n-k+1$ and in  the last step use Bernoulli's inequality).
As we are given that $k\ge 1$ and $n\ge k^3$, we have
$$ n\ge k^3>(k^2-1)(k-1)=(k+1)(k-1)^2,$$
so that
$$ 1-\frac{(k-1)^2}n>1-\frac1{k+1}$$
and ultimately
$$\begin{align}4(k+1)!{n-1\choose k-1}&=4(k+1)k\cdot \frac{(k-1)!{n-1\choose k-1}}{n^{k-1}}\cdot n^{k-1}\\&>4(k+1)k\left(1-\frac1{k+1}\right)n^{k-1}\\&=4k^2 n^{k-1}.\end{align}$$
