Fix a natural number $n \in \mathbb{N}$. We define $T^{-0} E : = E$ and the equality $T^{-1} E = E$ is included in the definition of a measure-preserving endomorphism, so it is left to establish the claim for $n > 1$. Since we define
\begin{equation}
T^{-n} : = \underbrace{T^{-1} T^{-1} \cdots T^{-1}}_{n \text{ times}}
\end{equation}
then, applying
R1973: Map composition is associative, we have the recurrence relation
\begin{equation}
\begin{split}
T^{-n} E
= T^{-n + 1 - 1} E
= T^{- n + 1} T^{-1} E
= T^{- n + 1} E
\end{split}
\end{equation}
By a repeated application of this recurrence relation, we have the equalities
\begin{equation}
\begin{split}
T^{-n} E & = T^{- n + 1} E \\
T^{-n + 1} E & = T^{- n + 2} E \\
T^{-n + 2} E & = T^{- n + 3} E \\
& \; \; \vdots \\
T^{-3} E & = T^{-2} E \\
T^{-2} E & = T^{-1} E \\
T^{-1} E & = E \\
\end{split}
\end{equation}
By transitivity of equality, the claim follows. $\square$
