Proving upper bound on a term involving binomial coefficients I am trying to show:

$$\binom{n}{k}\left(1-\frac{k}{n}\right)^{2n-2k+2} \left(\frac{k-1}{n}\right)^{2k}\leq \frac{1}{n^2}$$ for $k\in \{2,3,\cdots,n-1\}$ for all $n$

I observe that numerically that is true but analytically, I am not able to look at the term on LHS as a term of a binomial expansion or something like that to get an upper bound. Any ideas?

This is not a proof.
$$a_k=\binom{n}{k}\left(1-\frac{k}{n}\right)^{2n-2k+2} \left(\frac{k-1}{n}\right)^{2k}$$
Expanding as series for large $n$ and keeping the first term only, we have
$$a_2=\frac{1}{2 e^4 n^2}+O\left(\frac{1}{n^3}\right)\qquad \qquad
a_3=\frac{32}{3 e^6 n^3}+O\left(\frac{1}{n^5}\right)$$
$$a_4=\frac{2187}{8 e^8 n^4}+O\left(\frac{1}{n^5}\right)\qquad \qquad
a_5=\frac{131072}{15 e^{10} n^5}+O\left(\frac{1}{n^6}\right)$$
$$a_{n-4}=\frac{131072}{3 e^{10} n^6}+O\left(\frac{1}{n^7}\right)\qquad \qquad
a_{n-3}=\frac{2187}{2 e^8 n^5}+O\left(\frac{1}{n^6}\right)$$
$$a_{n-2}=\frac{32}{e^6 n^4}+O\left(\frac{1}{n^5}\right)\qquad \qquad
a_{n-1}=\frac{1}{e^4 n^3}+O\left(\frac{1}{n^5}\right)$$
