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This lacunary power series has radius of convergence one. For $|z|<1$, the inequality $n!\ge n$ for $n\ge 1$ gives absolute convergence by a geometric series comparison. For $|z|>1$ its terms do not tend to zero. At every point of the unit circle the terms have modulus one, so the term test for divergence proves divergence there too. The two initial terms both equal $z$, since $0!=1!=1$; this finite repetition does not affect any convergence conclusion.

For $|z|=r<1$, the inequality $n!\ge n$ for $n\ge 1$ givesThe first $r$ comes from $0!=1$; both initial terms have exponent one. Thus the factorial-gap power series converges absolutely inside the unit circle. For $|z|>1$, its terms have magnitudes $|z{|}^{n!}\to \infty$, so the term test for divergence proves divergence and $R=1$.

At every point with $|z|=1$, each term has magnitude one, so the terms cannot tend to zero, regardless of their phases. HenceThe large gaps between exponents make this a lacunary power series; cancellation cannot overcome the necessary condition that the terms of a convergent series tend to zero.

This is a lacunary power series: the coefficient of $z^{n!}$ is $n^n$, while the other coefficients are zero. Along its nonzero coefficients,The factorial dominates $n\log n$; for example $n!\ge n(n-1)(n-2)(n-3)$ for large $n$ already suffices for this limit. Hence the coefficient limsup in the Cauchy-Hadamard theorem is one. The radius of convergence is $\boxed{R=1}$.

Directly, for $|z|<1$ the consecutive term ratio is $(n+1{)}^{n+1}{n}^{-n}|z{|}^{nn!}$, which tends to zero; for $|z|\ge 1$ the term magnitudes are at least $n^n$ and fail to tend to zero. The factorial is the exponent of $z$, not part of its coefficient.