Faulhaber-like series The series
$$F(n,m) := 1^n + 2^n + \ldots + m^n = \sum_{k=1}^m k^n$$
Is known as the Faulhaber's Series.
I've tried to find a formula for this similar series but I've failed so far.
$$\mathcal{F}(n,m) := m^n - (m-1)^n + (m-2)^n -\ldots\pm 1^n= \sum_{k=1}^m (-1)^{m-k}k^n$$

EDIT:
Thanks to André Nicolas' answer I arrived to the following formula:
$$\mathcal{F}(n,m) = (-1)^m\left(F(n,m) - 2^{n+1}F\left(n,\lfloor m/2\rfloor\right)\right)$$
 A: Hint: For $1^4-2^4+3^4-4^4+5^4-6^4+7^4-8^4+9^4$ we would use 
$$(1^4+2^4+\cdots +9^4)-(2)(2^4+4^4+6^4+8^4).$$
And $2^4+4^4+6^4+8^4=2^4(1^4+2^4+3^4+4^4)$.
One can call it Faulhaber  minus Faulhaber.
A: The "generating-function hammer" approach I mentioned above proceeds by computing a formal power series in $\mathcal{F}_{n,m}$:
\begin{align}
\mathcal{F}_m(x)\equiv\sum_{n=0}^\infty \mathcal{F}_{n,m}\frac{x^n}{n!}
&=\sum_{n=0}^\infty\left(\sum_{k=1}^m (-1)^{m-k}k^n\right)\frac{x^n}{n!}\\
&=\sum_{k=1}^m(-1)^{m-k}\sum_{n=0}^\infty \frac{(kx)^n}{n!}\\
&=\sum_{k=1}^m(-1)^{m-k} e^{kx}\\
&=(-1)^m \sum_{k=1}^m (-e^x)^k \\&=(-1)^m \frac{(-e^x)-(-e^x)^{m+1}}{1-(-e^x)}=\frac{e^{mx}\pm 1}{1+e^{-x}}
\end{align}
where we have used the Taylor series for the exponential and the geometric series. (The $\pm$ in the last equation is for odd/even $m$.)
Now, suppose we had not done the alternating series. One may confirm easily that we would have instead received the generating function $$\mathcal{G}_m(x)=\sum_{k=1}^m e^{kx}=\frac{e^{mx}-1}{1-e^{-x}}$$ (Note that $\mathcal{G}_{n,m}$ is identical to the $F_{n,m}$ defined in the problem; it's merely a matter of personal taste that I use this instead.) Observe that 
\begin{align}
\mathcal{F}_m(x)\mp \, \mathcal{G}_m(x)
&=\frac{e^{mx}\pm 1}{1+e^{-x}}\mp \frac{e^{mx}-1}{1-e^{-x}} \\
&=-2\frac{e^{(m-1)x}-1}{1-e^{-2x}}\\
&=-2 G_{\frac{m-1}{2}}(2x), \hspace{1.8cm}(m \text{ odd})\\
&=+2\frac{e^{(m/2)x}-1}{1-e^{-2x}}\\
&=+2 G_{\frac{m}{2}}(2x), \hspace{2cm}(m \text{ even})\\
\end{align}
Identifying coefficients, we conclude that 
\begin{align}
\mathcal{F}_{n,m}
&=\begin{cases}
\mathcal{G}_{n,m}-2^{n+1}G_{n,\frac{m-1}{2}},& m\text{ odd}\\
-\mathcal{G}_{n,m}+2^{n+1}G_{n,\frac{m}{2}},& m\text{ even}
\end{cases}\\
&=(-1)^{m-1}\left(\mathcal{G}_{n,m}-2^{n+1}G_{n,\lfloor m/2 \rfloor}\right)
\end{align}
So we have derived (by much more formal means) the same equation glimpsed from Andre's answer. Note that while we had to struggle to relate this to the generating function for the Faulhaber result, getting to $\mathcal{F}(x)$ itself was rather simple. This is useful, since a singularity analysis of this generating function yields (via the Cauchy integral formula and steepest descent) asymptotic results for the alternating Faulhaber series. The curious reader should see Flajolet & Sedgwick's Analytic Combinatorics for a full treatment of this subject.



