I am reading Hatcher's treatment of the Adam's spectral sequence. http://pi.math.cornell.edu/~hatcher/SSAT/SSch2.pdf

On page 20, he states "Thus for each $i$ the groups $\pi_i(Z^k)$ are zero for all sufficiently large $k$. The same is true for the groups $\pi^Y_i (Z^k)$ when $Y$ is a finite spectrum, since a map $\Sigma^i Y \to Z^k$ can be homotoped to a constant map one cell at a time if all the groups $\pi_j (Z^k)$ vanish for $j$ less than or equal to the largest dimension of the cells of $\Sigma^i Y$.''

Context: $Z^k$ is a CW spectrum of finite type. Also, $\pi^Y_n(Z) = [ \Sigma^n Y, Z ] = colim_k [ \Sigma^{n} Y_k, Z_k]$, where the $Y_k$ and $Z_k$ are spaces.

I don't really understand the details of his statement. So my questions are

1) Can someone explain the details of this more?

2) (I'd prefer) Can we say instead that this holds because we can write $Y$ as a finite limit of sphere spectra, and finite limits and filtered colimits commute? And if so, how does one write $Y$ as a finite limit of sphere spectra?... Does this work exactly the same as if we had spaces?