Your expression is the difference of two central Eulerian numbers ,

$$A(k):=\sum_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2}{2k+1 \choose j}=\left \langle {2k-2\atop k-2} \right \rangle-\left \langle {2k-2\atop k-3} \right \rangle$$

as you can easily deduce from their closed formula. The positivity of $A(k)$ is just due to the fact that the Eulerian numbers $\left \langle {n\atop j}\right \rangle$ are increasing for $1\leq j\leq n/2$ (like the binomial coefficients); this fact has a clear combinatorial explanation also.

See e.g.

http://en.wikipedia.org/wiki/Eulerian_number

http://www.oeis.org/A008292

**[edit]**: although by now all details have been very clearly explained by Victor Protsak, I wish to add a general remark, should you find yourself in an analogous situation again. A healthy approach in such cases is adding variables, following the motto "more variables = simpler dependence" (like when one passes from quadratic to bilinear). In the present case, you may consider

$$A(k):=a(k,\, , 2k-2,\, ,2k+1)$$

where you define

$$a(k,n,m):=\sum_{j=0}^{k-1}(-1)^{j+1}(k-j)^{n}{m \choose j}$$

in which it is more apparent the action of the iterated difference operator, or, in the formalism of generating series, the Cauchy product structure:

$$\sum_{k=0}^\infty a(k,n,m)x^k=-\sum_{j=0}^\infty j^nx^j\, \sum_{j=0}^\infty(-1)^j{m \choose j} x^j =-(1-x)^m\sum_{j=0}^\infty j^nx^j. $$

The series
$$\sum_{j=0}^\infty j^nx^j$$
is now quite a simpler object to investigate, and in fact it is well-known to whoever played with power series in childhood. It sums to a rational function

$$(1-x)^{-n-1}x\sum_{k=0}^{n}\left \langle {n\atop k}\right \rangle x^k$$
that defines the Eulerian polynomial of order $n$ as numerator, and the Eulerian numbers as coefficients. In your case, $m=n+3$, meaning that you are still applying a discrete difference twice (in fact just once, due to the symmetric relations; check Victor's answer).