The scalar and matrix phi series converge absolutely

Put $R:=\mathrm{\mid }z\mathrm{\mid }$. Since $(j+k)!\ge j!$ for every $j\ge 0$, $\mathrm{\mid }{z}^j\mathrm{/}(j+k)!\mathrm{\mid }\le R^j\mathrm{/}j!$; the majorant sums to $e^R$ by [F4]. Thus the scalar series is absolutely convergent.

If $n=0$, the series definition gives $X^j=I_0=0_{0\times 0}$ for every $j\ge 0$. Every term has norm $0$ by [F1], so the matrix series is absolutely convergent and every partial sum equals the unique empty matrix. For $n\ge 1$, put $R:=\mathrm{\parallel }X{\mathrm{\parallel }}_2$. The operator norm is defined by [F1]. Repeated submultiplicativity in [F2] gives $\mathrm{\parallel }{X}^j{\mathrm{\parallel }}_2\le R^j$, so $\mathrm{\parallel }{X}^j\mathrm{/}(j+k)!{\mathrm{\parallel }}_2\le R^j\mathrm{/}j!$ for every $j$. The norms therefore have a finite sum by [F4].

For $n\ge 1$, let $S_N$ be the partial sums of the matrix series. For $N>M$, the triangle inequality bounds $\mathrm{\parallel }{S}_N-S_M{\mathrm{\parallel }}_2$ by the tail of the convergent scalar majorant in step 1.2, so $(S_N)$ is Cauchy. The matrix space is finite-dimensional and complete by [F3], hence $(S_N)$ converges in operator norm. For $n=0$, convergence was established directly in step 1.2.

At $z=0$ only the $j=0$ term of the scalar series is nonzero, and at $X=0$ with $n\ge 1$ only the $j=0$ matrix term is nonzero; these terms are respectively $1\mathrm{/}k!$ and $I_n\mathrm{/}k!$. When $n=0$, every matrix term and $I_0\mathrm{/}k!$ is the unique empty matrix by step 1.2. Hence both stated zero values follow for every allowed size.