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$x^2-ax-(a^2+1)x+a(a^2+1)=0$ $\implies (x-a)(x-(a^2+1))=0$

Clearly greater root is $a^2+1=\alpha\text{(Let)}.~~$ Let $f(x)=x^2-a^2x-2(a^2-2)=0$ . Then the condition that $\alpha$ lies between the roots of $f(x)$ is $f(\alpha)<0$ . Solving we get $a^2>5$ from where we get least positive integral value of $a$ as $3$ .

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• $\sqrt{4\sqrt{4\sqrt{4\sqrt{\cdots}}}}=4^{1/2+1/4+1/8+\cdots}=4$
• $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}=2$
• $\sqrt 5\sqrt{\sqrt 5\sqrt{\sqrt 5\sqrt{\sqrt 5\sqrt{\cdots}}}}=\sqrt 5^{1+1/2+1/4+\cdots}=5$
• $\sqrt{\dfrac{27}{\sqrt{\dfrac{27}{\sqrt{\dfrac{27}{\sqrt{\cdots}}}}}}}=27^{1/2-1/4+1/8+\cdots}=\textbf{3}$

Therefore answer is $\boxed{\sqrt{\dfrac{27}{\sqrt{\dfrac{27}{\sqrt{\dfrac{27}{\sqrt{\cdots}}}}}}}}$ .