Linear Map of nilpotent endomorphism is an automorphism [closed]

I struggle a lot about this proof: Assume a nilpotent endomorphism $\psi: W \rightarrow W$, than the mapping has to be an automorphism.

From the definition I have to show, that the the mapping is linear and bijective. I also know that nilpotent means: $ \exists n \in \mathbb{N} \ s. t. \ \psi ^{n}=0 $.

But how i can this proof?

This is never true. Nilpotent implies not surjective: Indeed, if we assume surjective, then $\psi^{m}$ is surjective as well, and we see that $$\text{ker}(\psi^n) = \lbrace w\in W \: \mid \psi(\psi^{n-1}(w)) = 0\rbrace = \lbrace w\in \psi^{n-1}(W) \: \mid \psi(w) = 0\rbrace = \text{ker}(\psi) \: ,$$ so the kernel stays the same, but $\text{ker}(\psi^n)=\text{ker}(0)=W$. Thus, it must be the zero map, which is not surjective, a contradiction.

As a note, nilpotent also implies not injective.

Edit: If you meant that $1+\psi$ is an automorphism, you can use that $$ (1+\psi)(1-\psi+\psi^2-\dots +(-1)^{n-1}\psi^{n-1}) = 1\pm \psi^n = 1$$ to get an inverse, therefore proving bijectivity. Linearity follows directly.