Prove $f(A)\in Mat_{n \times n}$ is invertible 
Let $A\in Mat_{n\times n}(F)$ be a nilpotent matrix, and $f\in F[x]$ a polynomial such that $f(0)\neq 0$ Prove: $f(A)\in Mat_{n \times n}$ is invertible 

I have started with let assume that $f(A)\in Mat_{n \times n}$ is  not invertible $\Rightarrow$ there is no free variable in the minimal polynomial $\Rightarrow$ $0$ is a root of each polynomial has the minimal polynomial divides each $f(A)=0$ in therefore $f(0)=0$ and that is not true
I know that the proof does not flow correctly any ideas? 
Due to previous answers I am looking for a proof without Rings
Second try: 
$f(0)\neq 0$ $Rightarrow$ W.L.G $f(x)=x^2+x+1$ in general a polynomial with a free variable, therefore $f(A)=A^2+A+1\Rightarrow A$ is upper triangular matrix therefore the determinant$\neq 0$
 A: Hint: you can write $f(A)=f(0)I+AB$ for some matrix $B$. 
A: Show first that if $N$ is a nilpotent matrix then $I + N$ is invertible. This can be done by guessing a formal inverse
$$ \frac{1}{I + N} = \sum_{k=0}^{\infty} (-N)^k = \sum_{k=0}^n (-N)^k $$
and verifying that this indeed works. Slightly more generally, if $a \neq 0$ then $aI + N = a \left(I + \frac{N}{a} \right)$ is also invertible with inverse $\frac{1}{a} \left(I + \frac{N}{a} \right)^{-1} = \sum_{k=0}^n \frac{(-1)^k}{a^{k+1}} N^k.$
Now, writing $f(x) = a_0 + a_1 x + \dots + a_m x^m$, we have
$$ f(A) = a_0 I + a_1 A + \dots + a_m A^m = a_0 I + B $$
where $B$ is defined by $B = a_1 A + \dots + a_m A^m$ and $a_0 = f(0)$. The binomial theorem can be used to show that the sum of any number of commuting nilpotent matrices is itself nilpotent and so $B$ is nilpotent and $f(A)$ is invertible by the discussion above.

Alternatively, $A$ is similar to a strictly upper triangular matrix (this is a weak form of the Jordan form of nilpotent matrices) and so $f(A)$ is a matrix that is similar to an upper triangular matrix whose diagonal elements are $f(0) \neq 0$ which is an invertible matrix.
