Expected number of times a set of 10 integers (selected from 1-100) is selected before all 100 are seen

Suppose I have a set of 100 integers. I randomly choose 10 of those, make a note of which ones I selected, and repeat the process. What is the expected number of times this process must be repeated before I see all 100 integers?

I'm also greatly interested in how this is calculated as I'm trying to increase the expected number of times this process is repeated by changing the set sizes.

As pointed out in a comment, this is the coupon collector's problem with coupons collected in groups (with distinct coupons per group). Let $n$ be the number of coupons and $m$ the number of coupons drawn per group. Denote the number of steps required to draw all coupons by $T$. By inclusion-exclusion,

$$ P(T\le t)=\sum_{j=0}^n(-1)^j\binom nj\left(\frac{\binom{n-j}m}{\binom nm}\right)^t\;. $$

Thus the expected number of steps required is

\begin{align} E[T]&=\sum_{t=0}^\infty P(T\gt t) \\ &= \sum_{t=0}^\infty\sum_{j=1}^n(-1)^{j-1}\binom nj\left(\frac{\binom{n-j}m}{\binom nm}\right)^t \\ &= \sum_{j=1}^n(-1)^{j-1}\binom nj\sum_{t=0}^\infty\left(\frac{\binom{n-j}m}{\binom nm}\right)^t \\ &= \sum_{j=1}^n(-1)^{j-1}\binom nj\frac1{1-\frac{\binom{n-j}m}{\binom nm}}\;. \end{align}

In your case $n=100$ and $m=10$, so

\begin{align} \sum_{j=1}^{100}(-1)^{j-1}\binom{100}j\frac1{1-\frac{\binom{100-j}{10}}{\binom{100}{10}}} \approx49.945 \end{align}

groups have to be drawn on average, corresponding to about $49.945$ groups of 10 coupons, compared to the $100H_{100}\approx518.74$ individual coupons required on average without groups.