Relative homology of skeleta of CW-complexes I am trying to show that, if $(X, A)$ is a relative CW-complex where $X^p$ denotes the $p$-skeleton, then $H_n(X^p, X^{p-1})$ is zero when $n\neq p$ and equal to $\bigoplus_\alpha \mathbb{Z}$, where $\alpha$ is the number of $p$-cells of $X$, when $n=p$. 
For this, I try to show that $H_n(X^p, X^{p-1})\cong H_n(\coprod_\alpha D^p, \coprod_\alpha S^{p-1})$. Since I am working only with CW-complexes of finite type, I tried to do some induction on the number of $p$-cells of $X$, but it does not work very well. Could you help me? 
Warning: I don't want to use the fact that $X^p/X^{p-1}\cong \bigvee_\alpha S^p$ and then use $H_n(X^p, X^{p-1})\cong \widetilde{H}_n(X^p/X^{p-1})$. 
Thank you for your help!

Write $B_r=\{x\in\mathbb{R}^p:|x|<r\}$ and let $f_\alpha:\overline{B_1}\to X^p$ be the inclusion of the $\alpha$th $p$-cell of $X$.  Let $$Y=X^{p-1}\cup\bigcup_\alpha f_\alpha(\overline{B_1}\setminus B_{1/2})\subseteq X^p.$$  Note that $Y$ deformation-retracts onto $X^{p-1}$ by radially projecting to the boundary in each $p$-cell.  Thus we have an isomorphism $$H_*(X^p,X^{p-1})\cong H_*(X^p,Y).$$  Furthermore, $X^{p-1}$ is closed and contained in the interior of $Y$, so by excision, $$H_*(X^p,Y)\cong H_*(X^p\setminus X^{p-1}, Y\setminus X^{p-1}).$$  But the pair $(X^p\setminus X^{p-1},Y\setminus X^{p-1})$ is homeomorphic to $(\coprod_\alpha B_1,\coprod_\alpha B_1\setminus B_{1/2})$.  This deformation-retracts to $(\coprod_\alpha \overline{B_{1/2}}, \coprod_\alpha \partial B_{1/2})\cong(\coprod_\alpha D^p,\coprod_\alpha S^{p-1})$, and the desired result follows.
