The coefficients of an Euler-Frobenius polynomial is given by
$$b_k^n=\sum_{\ell=1}^k(-1)^{k-\ell}\ell^n\left(\begin{matrix}n+1\\k-\ell\end{matrix}\right)$$
The symmetry property of this coefficients say that $b^n_k=b^n_{n+1-k}$
I've tried quite a few things but haven't been able to prove that equality neither find much information about these coefficients. I was looking for ideas that could lead me towards the proof.
Suppose we have the coefficient of the Euler-Frobenius polynomial $$b_k^n = \sum_{l=1}^k (-1)^{k-l} l^n {n+1\choose k-l}$$ and we seek to show that $b_k^n = b_{n+1-k}^n$ where $0\le k\le n+1.$
First re-write this as $$\sum_{l=0}^{k} (-1)^{l} (k-l)^n {n+1\choose l}.$$
Introduce the Iverson bracket $$[[0\le l\le k]] = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{z^l}{z^{k+1}} \frac{1}{1-z} \; dz$$
and the exponentiation integral $$(k-l)^n = \frac{n!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} \exp((k-l)w) \; dw.$$
to get for the sum (extend the summation to $n+1$ since the Iverson bracket controls the range)
$$\frac{n!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} \exp(kw) \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+1}} \frac{1}{1-z} \sum_{l=0}^{n+1} {n+1\choose l} (-1)^l z^l \exp(-lw) \; dz \; dw \\ = \frac{n!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} \exp(kw) \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+1}} \frac{1}{1-z} (1-z\exp(-w))^{n+1} \; dz \; dw.$$
Evaluate this using the residues at the poles at $z=1$ and at infinity.
We obtain for $z=1$ $$-\frac{n!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} \exp(kw) (1-\exp(-w))^{n+1} \; dw,$$
note however that $1-\exp(-w)$ starts at $w$ so the power starts at $w^{n+1}$ making for a zero contribution.
We get for the residue at infinity $$-\mathrm{Res}_{z=0} \frac{1}{z^2} z^{k+1} \frac{1}{1-1/z} (1-\exp(-w)/z)^{n+1} \\ = -\mathrm{Res}_{z=0} z^{k} \frac{1}{z-1} (1-\exp(-w)/z)^{n+1} \\ = \mathrm{Res}_{z=0} \frac{z^{k}}{z^{n+1}} \frac{1}{1-z} (z-\exp(-w))^{n+1}.$$
We need to flip the sign on this one more time since we are exploiting the fact that the residues at the three poles sum to zero. Actually extracting the coefficient we get $$-\sum_{q=0}^{n-k} {n+1\choose q} (-1)^{n+1-q} \exp(-(n+1-q)w).$$
Substitute this into the integral in $w$ to get $$-\sum_{q=0}^{n-k} {n+1\choose q} \frac{n!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} \exp(kw) (-1)^{n+1-q} \exp(-(n+1-q)w) \; dw \\ = -\sum_{q=0}^{n-k} {n+1\choose q} (-1)^{n+1-q} (-1)^n (n+1-k-q)^n \\ = \sum_{q=0}^{n-k} {n+1\choose q} (-1)^q (n+1-k-q)^n.$$
Using the fact that $n+1-k-q$ is zero at $q=n+1-k$ we finally obtain $$\sum_{q=0}^{n+1-k} {n+1\choose q} (-1)^q (n+1-k-q)^n$$ which is precisely $b_{n+1-k}^n$ by definition, QED.
Addendum. An alternate proof (variation on the theme from above) starts from the unmodified definition and introduces
$${n+1\choose k-l} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k-l+1}} (1+z)^{n+1} \; dz.$$
This controls the range so we may extend $l$ to infinity. Introduce furthermore $$l^n = \frac{n!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} \exp(lw) \; dw.$$
These two yield for the sum $$\frac{n!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(-1)^k}{z^{k+1}} (1+z)^{n+1} \sum_{l\ge 0} (-1)^l z^l \exp(lw) \; dz \; dw \\ = \frac{n!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(-1)^k}{z^{k+1}} (1+z)^{n+1} \frac{1}{1+z\exp(w)} \; dz \; dw \\ = \frac{n!}{2\pi i} \int_{|w|=\epsilon} \frac{\exp(-w)}{w^{n+1}} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(-1)^k}{z^{k+1}} (1+z)^{n+1} \frac{1}{z+\exp(-w)} \; dz \; dw.$$
We evaluate this using the negatives of the residues at $z=-\exp(-w)$ and at infinity. We get for $z=-\exp(-w)$
$$\frac{n!}{2\pi i} \int_{|w|=\epsilon} \frac{\exp(-w)}{w^{n+1}} \frac{(-1)^k}{(-1)^{k+1} \exp(-(k+1)w)} (1-\exp(-w))^{n+1} \; dw \\ = - \frac{n!}{2\pi i} \int_{|w|=\epsilon} \frac{\exp(kw)}{w^{n+1}} (1-\exp(-w))^{n+1} \; dw.$$
As before the exponentiated term starts at $w^{n+1}$ so there is no coefficient on $w^n$ for a contribution of zero.
We get for the residue at infinity (starting from the next-to-last version of the integral)
$$-\mathrm{Res}_{z=0} \frac{1}{z^2} (-1)^k z^{k+1} \frac{(1+z)^{n+1}}{z^{n+1}} \frac{1}{1+\exp(w)/z} \\ = -\mathrm{Res}_{z=0} \frac{1}{z^2} (-1)^k z^{k+1} \frac{(1+z)^{n+1}}{z^{n+1}} \frac{z/\exp(w)}{1+z/\exp(w)} \\ = -\mathrm{Res}_{z=0} (-1)^k z^{k} \frac{(1+z)^{n+1}}{z^{n+1}} \frac{\exp(-w)}{1+z/\exp(w)}.$$
Doing the sign flip and simplifying we obtain $$\exp(-w) (-1)^k \times \mathrm{Res}_{z=0} \frac{(1+z)^{n+1}}{z^{n-k+1}} \frac{1}{1+z/\exp(w)}.$$
Extract the residue to get $$\exp(-w) (-1)^k \sum_{q=0}^{n-k} {n+1\choose q} (-1)^{n-k-q} \exp(-(n-k-q)w)$$
Substitute into the integral in $w$ to obtain $$\sum_{q=0}^{n-k} {n+1\choose q} \frac{n!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} (-1)^{n-q} \exp(-(n+1-k-q)w) \; dw \\ = \sum_{q=0}^{n-k} {n+1\choose q} (-1)^{n-q} (-1)^{n} (n+1-k-q)^n \\ = \sum_{q=0}^{n-k} {n+1\choose q} (-1)^q (n+1-k-q)^n.$$
We have obtained $b_{n+1-k}^n$ as before.
