Do the coefficients of these polynomials alternate in sign?

I reckon that $$p_{2n-1}(x)=\sum_{k=0}^{n-1}(-1)^k \binom{n+1-k}k x^{n-1-k}$$ and $$p_{2n}(x)=\sum_{k=0}^{n-1}(-1)^k \binom{n+2-k}k x^{n-1-k}$$ and that these can be proved by induction.

Lord Shark's answer is good (but might contain some subtle mistakes) ... I reckon ... \begin{eqnarray*} p_{2i}(x) = \sum_{j=0}^{i-1} \binom{2i-j-1}{j} (-1)^j x^{i-j-1} \\ p_{2i-1}(x) = \sum_{j=0}^{i-1} \binom{2i-j-2}{j} (-1)^j x^{i-j-1}. \\ \end{eqnarray*} Proof of the even formula ... \begin{eqnarray*} &p_{2i-1}(x) -p_{2i-2}(x) =\\ & x^{i-1}+ \sum_{j=0}^{i-2} \binom{2i-j-3}{j+1} (-1)^{j+1} x^{i-j-2} - \sum_{j=0}^{i-1} \binom{2i-j-3}{j} (-1)^j x^{i-j-2} \\ &= x^{i-1}+ \sum_{j=0}^{i-2} \binom{2i-j-2}{j} (-1)^{j+1} x^{i-j-2} = p_{2i}(x). \end{eqnarray*} And the answer to the question in the title is $\color{red}{\text{yes}}$.