Exponential of a rank-one operator ($e^{tK}=I+\frac{e^{t\phi (a)}-1}{\phi (a)}K$)

For the bounded rank-one operator $Kx=a\phi (x)$, insert the powers of a rank-one operator into the convergent operator exponential power series. Writing $c=\phi (a)$ gives the displayed formula for $c\ne 0$. At $c=0$ the quotient has limit $t$, so $e^{tK}=I+tK$. This includes the zero operator and the nonzero nilpotent operator cases without dividing by zero.