Given 315, find
$$
S=\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+\cdots+\sqrt[2011]{2011}}}}
$$

the integer part of \( S \).
Let $n=2011$, and set
$$
\begin{array}{l}
a_{1}=\sqrt[n]{n}, \\
a_{k}=\sqrt[n-k+1]{n-k+1+a_{k-1}}(2 \leqslant k \leqslant n-1) .
\end{array}
$$

Then $S=a_{n-1}$.
We will prove by induction that:
$$
1<a_{m}<2(m=1,2, \cdots, n-1) \text {. }
$$
(1) When $m=1$, since $1<n<2^{n}$, we have
$$
1<\sqrt[n]{n}<2 \text {. }
$$
(2) Assume that when $m=k(1 \leqslant k \leqslant n-2)$, the conclusion holds. Then when $m=k+1$, since
$$
\begin{array}{l}
1<n-k+1<n-k+a_{k}<n-k+2 \\
\leqslant(1+1)^{n-k}=1+(n-k)+\cdots+1,
\end{array}
$$

we have $1<a_{k+1}=\sqrt[n-k]{n-k+a_{k}}<2$.
Therefore, when $m=k+1$, the conclusion holds.
In summary, $1<a_{m}<2(m=1,2, \cdots, n-1)$.
Thus, the integer part of $S$ is 1.