If $1$ is an eigenvalue, then if we let $v$ be an eigenvector for $1$, we get that $(A^{k-1} + \dots + A + I)v = kv \neq 0$
If $1$ is not an eigenvalue, then $A-I$ is invertible, so we can multiply the equation $0=(A-I)(A^{k-1} + \dots + A + I)$ with $(A-I)^{-1}$