Prove that $T = I$ Problem. Suppose that $T^2 = I$ and suppose that $-1$ is not an eigenvalue of T. Prove that $T=I$. $\DeclareMathOperator{null}{Null}$

My attempt:
Suppose $v \in V$, and consider the vector $Tv-v$. If $Tv-v$ is an eigenvector, then
\begin{align}
T(Tv-v)&=\lambda(Tv-v) \\ 
T^2v-Tv &= \\ 
Iv-Tv &=\\
v-Tv &=\\
-Tv+v &=\\
-1(Tv-v) &=
\end{align}
so $\lambda = -1$. This is a contradiction to our assumption since $-1$ is not an eigenvalue. Thus, 
\begin{align}
\lambda &= 1 \\
&=1(Tv-v).
\end{align}
Now the eigenspace of $T$ is 
\begin{align}
\null(T-\lambda I) &=\null(T-I) \\ 
T-I &= 0 \\
T&=I
\end{align}
Am I correct with my definitions and assumptions?

I can't make any sense of your solution, sorry.
Hint: $T^2=I$ says $(T-I)(T+I)=0$. Now saying $-1$ is not an eigenvalue means _____. Hence _____, so $T-I=0$.
(Hint regarding the hint: The definition of "$-1$ is not an eigenvalue of $T$" has nothing to do with determinants. Look back - the definition was what?)
