= Solution
Work in the finite-dimensional <vector space> $V$ implicit in the requested matrix representation and $n=\dim V$. Put $K_j=\ker\alpha^j$, with $K_0=0$. These form an increasing sequence ending in $K_m=V$, and $\alpha(K_j)\subseteq K_{j-1}$. Choose a <basis> of $K_1$, extend it to a basis of $K_2$, and continue until a basis of $V$ is obtained. Each basis vector introduced at stage $j$ maps into the span of vectors chosen at earlier stages. Its matrix column can therefore have nonzero entries only in earlier rows. This proves that the matrix is \b[strictly upper triangular].
For a strictly upper-triangular $n\times n$ <matrix> $N$, a nonzero contribution to $(N^r)_{ij}$ requires an index chain $i<i_1<\cdots<i_{r-1}<j$. No such chain has $n$ links among $n$ indices. Consequently $\boxed{\alpha^n=0}$. An example with exact index four is the <Nilpotent Jordan block>
$$
\boxed{M=\begin{pmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{pmatrix}.}
$$
Here $M^3$ has entry one in position $(1,4)$ and all others zero, while $M^4=0$.
In the adapted basis, $I+A$ is upper triangular with diagonal entries one, so all its <eigenvalues> are one. This does not persist for the product of two such matrices with unrelated adapted bases. Take
$$
A=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad B=\begin{pmatrix}0&0\\1&0\end{pmatrix}.
$$
Both square to zero, but
$$
(I+A)(I+B)=\begin{pmatrix}2&1\\1&1\end{pmatrix}
$$
has <characteristic polynomial> $t^2-3t+1$ and <eigenvalues> $(3\pm\sqrt5)/2$, neither equal to one. The obstruction is that the two nilpotent matrices need not admit a common triangularizing basis.
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