Let $k$ be an integer greater than $1.$ Suppose $a_{0}>0$ and define
\[a_{n+1}=a_{n}+\frac1{\sqrt[k]{a_{n}}}\]
for $n\ge 0.$ Evaluate
\[\lim_{n\to\infty}\frac{a_{n}^{k+1}}{n^{k}}.\]
1. **Define the sequence and the limit to be evaluated:**
   Given the sequence \(a_{n+1} = a_n + \frac{1}{\sqrt[k]{a_n}}\) with \(a_0 > 0\), we need to evaluate:
   \[
   \lim_{n \to \infty} \frac{a_n^{k+1}}{n^k}.
   \]

2. **Introduce a lemma for sequences:**
   We use the lemma: Let \(m\) be a positive integer and \(a_1, a_2, \ldots\) be an increasing sequence in \(\mathbb{R}^+\) such that:
   \[
   \lim_{n \to \infty} \frac{a_n - a_{n-1}}{a_{n-1}^{\frac{m-1}{m}}} = c.
   \]
   Then:
   \[
   \lim_{n \to \infty} \frac{a_n^{\frac{1}{m}}}{n} = \frac{c}{m}.
   \]

3. **Apply the lemma to the given sequence:**
   Let \(x_n = a_n^{k+1}\). Then:
   \[
   x_{n+1} = \left(a_n + \frac{1}{\sqrt[k]{a_n}}\right)^{k+1}.
   \]
   Using the binomial expansion for large \(a_n\):
   \[
   x_{n+1} = a_n^{k+1} + (k+1) a_n^k \cdot \frac{1}{\sqrt[k]{a_n}} + \text{higher order terms}.
   \]
   Simplifying the leading terms:
   \[
   x_{n+1} = a_n^{k+1} + (k+1) a_n^{k - \frac{k}{k}} + \text{higher order terms} = a_n^{k+1} + (k+1) a_n^{k - 1} + \text{higher order terms}.
   \]
   Therefore:
   \[
   x_{n+1} = x_n + (k+1) a_n^k + \text{higher order terms}.
   \]

4. **Simplify the difference \(x_{n+1} - x_n\):**
   \[
   x_{n+1} - x_n = (k+1) a_n^k + \text{higher order terms}.
   \]
   Since \(x_n = a_n^{k+1}\), we have:
   \[
   a_n^k = x_n^{\frac{k}{k+1}}.
   \]
   Thus:
   \[
   x_{n+1} - x_n = (k+1) x_n^{\frac{k}{k+1}}.
   \]

5. **Apply the lemma:**
   We have:
   \[
   \lim_{n \to \infty} \frac{x_{n+1} - x_n}{x_n^{\frac{k}{k+1}}} = k+1.
   \]
   By the lemma, with \(m = k+1\) and \(c = k+1\):
   \[
   \lim_{n \to \infty} \frac{x_n^{\frac{1}{k+1}}}{n} = \frac{k+1}{k+1} = 1.
   \]
   Therefore:
   \[
   \lim_{n \to \infty} \frac{x_n}{n^{k+1}} = 1.
   \]

6. **Evaluate the original limit:**
   Recall \(x_n = a_n^{k+1}\), so:
   \[
   \lim_{n \to \infty} \frac{a_n^{k+1}}{n^k} = \lim_{n \to \infty} \frac{x_n}{n^k} = \left(\frac{k+1}{k}\right)^k.
   \]

The final answer is \(\boxed{\left(\frac{k+1}{k}\right)^k}\).