This is basically the same process that Plato used in another thread.

$$\sqrt[n]{n^{\sqrt{n}}} = n^{1/\sqrt{n}}$$

Since $f(x) = x^{1/\sqrt{x}}$ is continuous for $x>0$, we can consider the subsequence when $n$ is a perfect square:

Let $x_n=n^{1/(2n)}-1$. Then $(x_n+1)^{2n} = n$.

When $n\ge 1$:

$$\dfrac{n(n-1)}{2}x_n \le n \Longrightarrow x_n \le \dfrac{2}{n-1}$$

Since $\displaystyle \lim_{n \to \infty} x_n = 0$, we have $\lim_{n \to \infty} \sqrt[n]{n^{\sqrt{n}}} = 1$.