Show that if f is analytic in $|z|\leq 1$, there must be some positive integer n such that $f(\frac{1}{n})\neq \frac{1}{n+1}$

Show that if f is analytic in $|z|\leq 1$, there must be some positive integer n such that $f(\frac{1}{n})\neq \frac{1}{n+1}$

MY SOLUTION

If $f(\frac{1}{n})=\frac{1}{n+1}$, then for all points $z_{n}=\frac{1}{n}$ ,
$f(z_{n})=\frac{1}{\frac{1}{z_{n}}+1}$
or $f(z_{n})=\frac{z_{n}}{1 + z_{n}}$ Because ${z_{n}}$ has an accumulation point at 0, this implies that $f(z)=\frac{z}{1+z}$ throughout its domain of analyticity which yields a contradiction since f was assumed analytic at
$z=-1$.

Can anyone help me improve it?

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