If $T(p(t)) = p(t+1)$ then find its minimal polynomial where $T$ is a linear operator from $Bbb{P_n} rightarrow Bbb{P_n}$

Consider the basis of polynomials ${1,t,t^2,ldots,t^n}$. Then, from$T(t^k)=(t+1)^k=1+kt+binom{k}{2}t^2+cdots+1$, the matrix of $T$ with respect to this basis is $$begin{pmatrix}1&1&1&1&cdots&&cr 0&1&2&3&cdots&&cr 0&0&1&3&cdots&&cr0&0&0&1&&cr vdots&&&ddots&ddots&cr0&cdots&&&0&1end{pmatrix}$$

Since the matrix is triangular, all the eigenvalues are $1$ and its minimal polynomial is of the type $(I-A)^k$. In fact, since none of the terms of the supra-diagonal is zero, the minimum value of $k$ that makes it zero is $n$, $(I-A)^n=0$.

If $p ne 0$ then from $p(t+1) = lambda p(t)$ by comparing leading coefficients we get that necessarily $lambda = 1$. Therefore the only eigenvalue of $T$ is $1$ so the minimal polynomial must be of the form $(T-I)^k$. Indeed by Cayley-Hamilton certainly $(T-I)^{n+1}=0$ since $dim mathbb{P}_n=n+1$. Furthermore, for $1 le k le n$ and the polynomial $p(t) = t^k$ we have $$((T-I)^kp)(t) = sum_{j=0}^k {k choose j} (T^jp)(t) = sum_{j=0}^k {k choose j} p(t+j) = sum_{j=0}^k {k choose j} (t+j)^k$$ and for $t = 0$ it is $$((T-I)^kp)(t) = sum_{j=0}^k {k choose j} j^k > 0$$ so clearly $(T-I)^k ne 0$. Therefore the minimal polynomial has to be $(T-I)^{n+1}$.