Faulhaber's Formula to evaluate series

$$S = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k^{99}}{n^{100}} = \lim_{n \to \infty} \frac{1}{n^{100}} \cdot \overbrace{\sum_{k=1}^{n} k^{99}}^{\text{I}}$$

$$I = \sum_{k=1}^{n} k^{99} = \frac{1}{99} \cdot \sum_{j=0}^{99} (-1)^j \binom{100}{j} B_j n^{100 - j}$$

But that seems AWFULLY, difficult? What can be done?

Expanding Lucien's comment: $$ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k^{99}}{n^{100}} = \lim_{n \to \infty}\frac1n\sum_{k=1}^n\Bigl(\frac{k}{n}\Bigr)^{99}=\int_0^1x^{99}\,dx. $$

One doesn't need the full detail of Faulhaber's formula. The relevant fact is that $f_j(n)=\sum_{k=1}^n k^j$ is a polynomial in $n$ with leading coefficient $\dfrac{1}{j+1}$ and degree $j+1$. You can prove this by induction. Hence writing $f_j(n)=\dfrac{1}{j+1} n^{j+1}+O(n^j)$ gives that $\lim\limits_{n\to\infty}f_j(n)n^{-(j+1)}=\dfrac{1}{j+1}$.