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We attempt to verify for $n=2m$ that

$$\frac{n}{n+1}\frac{n^n-1}{(n-1)^{n+1}} = \sum_{k=1}^{n/2} {n-k\choose k-1} \frac{n^k}{(n-1)^{2k}}.$$

We obtain

$$\sum_{k=1}^{m} {2m-k\choose k-1} \frac{2^k m^k}{(2m-1)^{2k}} = \frac{2m}{(2m-1)^2} \sum_{k=0}^{m-1} {2m-1-k\choose k} \frac{2^k m^k}{(2m-1)^{2k}} \\ = \frac{2m}{(2m-1)^2} \sum_{k=0}^{m-1} {2m-1-k\choose 2m-1-2k} \frac{2^k m^k}{(2m-1)^{2k}} \\ = \frac{2m}{(2m-1)^2} \sum_{k=0}^{m-1} \frac{2^k m^k}{(2m-1)^{2k}} [z^{2m-1-2k}] (1+z)^{2m-1-k} \\ = \frac{2m}{(2m-1)^2} [z^{2m-1}] \sum_{k=0}^{m-1} \frac{2^k m^k}{(2m-1)^{2k}} z^{2k} (1+z)^{2m-1-k} \\ = \frac{2m}{(2m-1)^2} [z^{2m-1}] (1+z)^{2m-1} \frac{1-(2mz^2/(2m-1)^2/(1+z))^m}{1-2mz^2/(2m-1)^2/(1+z)} \\ = 2m [z^{2m-1}] (1+z)^{2m} \frac{1-(2mz^2/(2m-1)^2/(1+z))^m}{(2m-1)^2(1+z)-2mz^2} .$$

Observe that the second term in the numerator of the fraction is $z^{2m}$ times a term with no pole at zero which thus does not contribute to the coefficient on $[z^{2m-1}]$ and we are left with just

$$2m [z^{2m-1}] (1+z)^{2m} \frac{1}{(2m-1)^2(1+z)-2mz^2}$$

which is

$$2m \mathrm{Res}_{z=0} \frac{1}{z^{2m}} (1+z)^{2m} \frac{1}{(2m-1)^2(1+z)-2mz^2}.$$

Note that since $\lim_{R\to\infty} 2\pi R\times R^{2m}/R^{2m}/R^2 = 0$ the functional term has residue zero at infinity and we may evaluate the residue at zero with the negative of the two residues from the quadratic in the denominator of the fraction using the fact that the residues sum to zero. Re-writing for these residues we find

$$- \frac{1}{z^{2m}} (1+z)^{2m} \frac{1}{z+(2m-1)/(2m)}\frac{1}{z-(2m-1)} \\ = - \frac{1}{z^{n}} (1+z)^{n} \frac{1}{z-(1-n)/n}\frac{1}{z-(n-1)}.$$

We get for the first residue (flip sign)

$$\frac{n^n}{(1-n)^n} \frac{1}{n^n} \frac{1}{(1-n)/n-(n-1)} = \frac{n}{(1-n)^n} \frac{1}{1-n^2} \\ = (-1)^{n+1} \frac{n}{(n-1)^n} \frac{1}{n^2-1} = (-1)^{n+1} \frac{n}{(n-1)^{n+1}} \frac{1}{n+1} $$

and the second one

$$\frac{1}{(n-1)^n} n^n \frac{1}{n-1-(1-n)/n} = \frac{n^{n+1}}{(n-1)^n} \frac{1}{n^2-1} = \frac{n^{n+1}}{(n-1)^{n+1}} \frac{1}{n+1}.$$

We sum these two observing that with $n$ even $(-1)^{n+1}=-1$ to obtain

$$\frac{n^{n+1}}{(n-1)^{n+1}} \frac{1}{n+1} - \frac{n}{(n-1)^{n+1}} \frac{1}{n+1} = \frac{n}{n+1} \frac{1}{(n-1)^{n+1}} (n^n - 1)$$

which is what we sought to prove.