Assume that NP = DTIME(2^sqrt(n)), prove that DTIME(2^sqrt(n)) = DTIME(2^n)
It is enough to pad a special delimiter (say a comma) and $(|x|^2-|x|-1)$ 1's. Suppose $L_\mathrm{pad}= \left\{\left\langle x,1^{|x|^2-|x|-1} \right\rangle : x \in L\right\}$. Since $L\in\mathsf{DTIME}(2^n)$, there is a TM that can determine whether $x\in L$ in $O(2^{|x|})$ time. We then construct a new TM: given a string $y$, it first checks whether $y$ has the form $\left\langle x,1^{|x|^2-|x|-1} \right\rangle$, then simulates $M$ on $x$. This new TM determines whether $y\in L_\mathrm{pad}$ in $O(2^{|x|})=O\left(2^{\sqrt{|y|}}\right)$ time, so we have $L_{\mathrm{pad}}\in \mathsf{DTIME}\left(2^{\sqrt{n}}\right)$, thus $L_{\mathrm{pad}}\in\mathsf{NP}$ by your assumption. This means there exists a non-deterministic TM $M'$ that decides $L_{\mathrm{pad}}$ in polynomial time. Hence, we can construct a new non-deterministic polynomial-time TM that decides $L$ by first padding a comma and $(|x|^2-|x|-1)$ 1's to the input then simulating $M'$, which means $L\in\mathsf{NP}$.