Transitive closure preserves set-theoretic rank ($\mathrm{rank}\mathrm{TC}(x)=\mathrm{rank}x$)

Using the convention that $\mathrm{TC}(x)$ is the least transitive set containing $x$ as a subset, put $x_0=x$, $x_{n+1}=\bigcup x_n$ and $\mathrm{TC}(x)={\bigcup }_{n<\omega }{x}_n$. If $\mathrm{rank}x=\alpha$, every element encountered is an iterated member of an original element and has rank below $\alpha$. Thus the closure has rank at most $\alpha$, while $x\subseteq \mathrm{TC}(x)$ gives the reverse inequality. A convention requiring $x$ itself to be an element instead has rank $\alpha +1$.