commutes with a nilpotent matrix and invertible The exercise is:

Given $V$ a vector space, $\dim V<\infty$ and $T, N\in \mathcal{L}(V)$, where $N$ is nilpotent and $TN=NT$. Prove that: $T$ is invertible iff $T+N$ is invertible. Furthermore, $\det(T)=\det(T+N)$ and $p_T(t)=p_{T+N}(t)$.

I need a hint about how can I start this kind of exercise, which involves a nilpotent matrix.
 A: Hint: Note that $N$ and $T$ are simultaneously upper triangularizable.
A: For a rather different approach one can directly prove $p_T(t)=p_{T+N}(t)$ which then gives you the equality of determinants and in turn the equivalence of invertibility.  I assume $\dim V = m$ and assume the field has characterisitic zero (or if one wants to contemplate positive characteristic fields, then $\text{char}(\mathbb F)\gt m$)
lemma 1:
by nilpotence, $N^m = \mathbf 0$ and with commutativity this means you can apply the binomial theorem, so $\big(T+N\big)^j= T^j +\Big(\sum_{k=1}^{j-1} \binom{j}{k}T^kN^{j-k}\Big) + N^j$
lemma 2:
$T^k N^r$ for $r\geq 1$ is nilpotent, because
$(T^k N^r)^m =  T^{km} N^{rm}=  T^{km}\mathbf 0 = \mathbf 0$
and recall that since nilpotent matrices are similar to strictly upper triangular matrices, they necessarily have trace zero
main argument:
$\text{trace}\Big(T+N\Big)= \text{trace}\Big(T\Big)+\text{trace}\Big(N\Big)= \text{trace}\Big(T\big) + 0 $
and for $j\in\big\{1,2,...,m\big\}$
$\text{trace}\Big(\big(T+N\big)^j\Big)= \text{trace}\Big(T^j\Big) + \sum_{k=0}^{j-1} \binom{j}{k}\cdot\text{trace}\Big(T^kN^{j-k}\Big)= \text{trace}\Big(T^j\Big) + 0$
Thus by application of Newton's Identities $\Big(T+N\Big)$ and  $\Big(T\Big)$ have the same characteristic polynomial.
A: It suffices to prove that $T$ and $T+N$ have the same characteristic polynomial. The other two results follow immediately.
I shall give a proof that does not involve triangulation of $T$, because an attendee of a first course on linear algebra usually has not seen any proof of the existence of algebraic closure of an arbitrary field from any course.
Let $F$ be the underlying field and $x$ be an indeterminate. Denote by $F(x)$ be the field of fractions of $F[x]$. For ease of presentation we assume that $T$ and $N$ are matrices over $F$.
Since $\det(xI-T)$ is a monic polynomial in $x$, it is nonzero. Therefore $xI-T$ is invertible over $F(x)$. Then $(xI-T)^{-1}$ commutes with $N$. Hence $N_1=(xI-T)^{-1}N$ is nilpotent and it can be triangularised over $F(x)$ (we do not need $F(x)$ to be algebraically closed here because $0\in F(x)$). Therefore $\det(I-N_1)=1$ and in turn $\det(xI-T-N)=\det(xI-T)$.
