Proving simple bound on coupon collector
Divide the process into $n$ phases. The $m$th phase ($m=1,\ldots,n$) starts when the $(m-1)$st cell has been occupied, and ends when the $m$th cell has been occupied. The first phase takes time $1$. Suppose that the occupied cell is $i_1$. The second phase then takes expected time $1/(1-p_{i_1})$. Suppose that the occupied cell is $i_2$. The third phase then takes expected time $1/(1-p_{i_1}-p_{i_2})$. And so on. Overall, the expected time is
$$
1 + \frac{1}{1-p_{i_1}} + \frac{1}{1-p_{i_1}-p_{i_2}} + \cdots + \frac{1}{1-p_{i_1}-\cdots-p_{i_{n-1}}} = \\
\frac{1}{p_{i_1}+\cdots+p_{i_n}} + \frac{1}{p_{i_2}+\cdots+p_{i_n}} + \cdots + \frac{1}{p_{i_n}} \leq = \\
\frac{1}{np_1} + \frac{1}{(n-1)p_1} + \cdots + \frac{1}{p_1} = \\
\frac{H_n}{p_1} \leq \frac{\ln n + 1}{p_1}.
$$