Same z transformed function, but different answers of inverse z transform?
The equations do produce the same results. \$\small e(k)\$ is assumed unilateral, so the equations in method 2 should be qualified by \$\small k&gt;0\$, thus:

$$\small e(k)=\begin{cases}
   (-4)^{k-1}= -(1/4)(-4)^{k}, \:\:\:k&gt;0 &amp; \\\\
    0, \:\:\text{otherwise}\end{cases}$$

Also, remember that:

$$\small \delta (k)=\begin{cases}
   1, \:\:\:k=0 &amp; \\\\
    0, \:\:\text{otherwise}\end{cases}$$