Why the shift helps can be explained by considering the QR of $H-rI=QR$. Since H is unreduced Hessenberg, it's first $n-1$ columns are linearly independent for any $r$. But if $r=\lambda_1$, $H-rI$ is singular. Then $r_{nn}=0$ and the last row of $R$ is zero. Then the last row of $H'+rI=RQ$ is zero except $nn$ component is $r$. So $r=\lambda_1$ or close to $\lambda_1$ will make $H'+rI$ $n-1,n$ component zero or closer to zero. This explains why the shift help. But this explanation doesn't fit well with the power iteration, where the component of $x'=(A-rI)x$ in $v_1$ actually decreases.