If $p$ is a polynomial, then either $\,p(z)=z^n\,$ or $\,\max_{|z|=1}|p(z)|>1$.

Let us assume that $\lvert f(z)\rvert\le 1$, for all $\lvert z\rvert=1$, and show that in such case $f$ has to be equal to $z^n$.

Cauchy Integral formula implies that $$ n!=f^{(n)}(0)=\frac{n!}{2\pi i}\int_{|z|=1}\frac{f(z)\,dz}{z^{n+1}}=\frac{n!}{2\pi }\int_0^{2\pi} f(\mathrm{e}^{it})\,\mathrm{e}^{-int}\,dt, $$ and thus, $\int_0^{2\pi} f(\mathrm{e}^{it})\,\mathrm{e}^{-int}\,dt\in\mathbb R\,$ and $$ 1=\frac{1}{2\pi }\int_0^{2\pi} f(\mathrm{e}^{it})\,\mathrm{e}^{-int}\,dt\le \frac{1}{2\pi }\int_0^{2\pi} |f(\mathrm{e}^{it})|\,dt\le 1, \tag{1} $$ since $\lvert\, f(z)\rvert\le 1$, for all $\lvert z\rvert=1$. Therefore, $$ \mathrm{Re}\,\big(f(\mathrm{e}^{it})\, \mathrm{e}^{-int}\big)=|f(\mathrm{e}^{it})|=1, $$ for all $\,t\in[0,2\pi]$, and thus $$\,\mathrm{Im}\,\big(f(\mathrm{e}^{it})\,\mathrm{e}^{-int}\big)=0,$$ for all $\,t\in[0,2\pi]$. Consequently, $$ f(\mathrm{e}^{it})\,\mathrm{e}^{-int}=1,\quad\text{for all $t\in[0,2\pi]$,} $$ or equivalently $$ f(z)=z^n, $$ for $\lvert z\rvert=1$, which implies that $f(z)=z^n$.

Extension. From the proof above we can make the following generalization:

Let $f:U\to\mathbb C$ be analytic, where $U$ contains the closed unit disk. If $f^{(n)}(0)=n!$
then $$ \max_{|z|=1}|f(z)|>1, $$ unless $f(z)=z^n$.