Supose that W in a vector space with finite dimension. Let $f \in End(V)$, also $f^n \in End(V)$ for $ n $ in positive natural numbers. Where $f^0=Id$, and for $n\ge1$, $f=f^{n-1}\circ f$, $f^2=f$. Show that $dim (Span(\{f^n | n \in \Bbb N \}))\le dimV$.
Hint: use Cayley-Hamilton theorem to start an argument by induction
So for $dim(V)=1$ the characteristic polynomial gives $1-f=Id-f=0$ so $Id, f$ are linearly dependent $dim(Span\{f^i\})\lt2$ for $1\le i\le n$, since $Id(v)$ can not be $0$, $dim(Span\{f^n\})=1$.
am not sure what to do from there on, I tried the following but leads me no where.
For $dim(V)=n+1$ Caley-hamilton gives $f^{n+1}+ c_nf^n+ c_{n-1}f^{n-a}...+c_{1}f+c_0Id=0$ so $dim(Span\{f^i\})\lt n+1$ for $1\le i\le n+1$.
The induction is not on the dimension of $ V $.
You wrote and interpreted Cayley-Hamilton correctly, now use it to prove by induction $\underline{\text {on } N\geq n }$ that for any such $ N $, $$f^N\in\text {Span}(\{f^i\}_{i=0}^{n-1})$$