Define $\sigma_k(n) = \sum_{i=1}^n i^k$ then since $\sigma_1(n) = \frac{n(n+1)}{2}$
First to show that the coefficient of $n$ in $\sigma_{2k+1}$ is $b^{2k+1}$: It's immediate from applying the binomial theorem to $\sigma_{2k+1}(n) = \frac{(n+b)^{k+1}-b^{k+1}}{k+1}$. Then we can show this coefficient is always zero by showing these polynomials are divisible by something with $n^2$ as its lowest power.
This argument is due to Pascal (I could not possibly come up with it!) taken from here,
$\begin{array} . \sigma_1(n)^k &=& \sum_{m=1}^n \left[ \left(\frac{m(m+1)}{2}\right)^k - \left(\frac{(m-1)m}{2}\right)^k \right] \\ &=& \sum_{m=1}^n \sum_{r=0}^k \binom{k}{r} \left(\frac{m}{2}\right)^k m^r (1 - (-1)^{k-r}) \\ &=& \frac{1}{2^k} \sum_{r=0}^k \binom{k}{r} \sigma_{k+r}(n) (1 - (-1)^{k-r}) \end{array}$
Now consider $k$ odd, many of the terms cancel, we are left with the fact that $\sigma_1^k$ is a linear combination of various $\sigma_{i}$ for $i$ odd.
The relation $\sigma_3 = \sigma_1^2$ is classical and can be proved in a variety of simple ways. Induction on (odd) $k$ shows that every other odd one ($\sigma_5$, $\sigma_7$ and so on) is also divisible by $\sigma_1^2$. In fact you can try out these induction arguments concretely for given $k$ to express these summations in terms of triangular numbers. That is called producing the Faulhauber polynomials.
This proves that there is no $n$ terms in these odd ($\ge 3$) sums of powers, but since the coefficient of $n$ in $\sigma_{2k+1}$ is $b_{2k+1}$ this proves that they are all zero.