 Solution

No such entire function exists. If it did, the entire function $h(z{)}^2-z$ would vanish at the distinct points $(n/(n+1){)}^2$, whose accumulation point is one. The identity theorem would give $h(z{)}^2=z$ on all of $\mathbb{C}$.

At zero this forces $h(0)=0$. Differentiating the identity gives $2h(z)h^{\prime }(z)=1$, impossible at zero. Equivalently, a squared holomorphic function has an even-order zero, whereas $z$ has a simple zero. This is the obstruction expressed by no entire square root of the identity function; a square-root branch near one cannot extend holomorphically through zero.