We need to show that any two qubits of $\ket{GHZ_n} = \frac1{\sqrt 2} (\ket{00\ldots0} + \ket{11\ldots1})$ can be left in a maximally entangled state by a series of single qubit measurements. Obviously, measurement in the standard basis $\{\ket 0, \ket 1\}$ will not do, since the state is left unentangled after first measurement. But measurement in the basis $\{\ket{+} = \frac1{\sqrt2}(\ket{0} + \ket{1}),\text{ } \ket{-}= \frac1{\sqrt2}(\ket{0} - \ket{1})\}$ does the trick. To see this, we note that
(1)Repeated use of these two formulas will rewrite any two target qubits in terms of the maximally entangled Bell States $\ket{\Phi^+} = \frac1{\sqrt 2}(\ket{00} + \ket{11})$ and $\ket{\Phi^-} = \frac1{\sqrt 2}(\ket{00} - \ket{11})$, and the other qubits in terms of the $\{\ket{+}, \ket{-}\}$ basis. Here exemplified in the first two qubits for $n=4$ and $n=5$:
(2)In general,
(3)where $S_+,S_-$ are sums of two disjoint partitions of the set of $2^{n-2}$ basis elements for $n-2$ qubits:
(4)This demonstrates that measurement of the other qubits in the $\{\ket{+}, \ket{-}\}$ basis, will leave the two target qubits in one of the Bell states $\ket{\Phi^+}$ or $\ket{\Phi^-}$, both of which are maximally entangled.