TITLE: Upper bound of probability of not getting all values in independent draws
QUESTION [6 upvotes]: We draw $m$ uniform independent random values among $n$, with $m\ge n$. We consider the probability $p(m,n)$ that not all $n$ values have been drawn. We want an upper bound within a constant factor.
The probability that a given value was not drawn is $q(m,n)=(1-1/n)^m$. If these probabilities were independent, we could use $1-(1-q(m,n))^n$ for $p(m,n)$. However the probabilities are dependent and that expression yields dead wrong values.
How can we rigorously derive an upper bound? I needs not be tight, I'd be happy with something within any constant factor.

The question's motivation is to prove that $\displaystyle\lim_{n\to+\infty}p(n\left\lceil\log_2(n)\right\rceil,n)=0$, if possible with an explicit upper bound (towards answering an often-asked question about cryptographic hashes reaching their whole stated destination set, under a random oracle model).
This itself is plausible because (as studied in the coupon collector's problem) the expected number of draws to get all $n$ values is $n(\ln(n)+\gamma)+\frac12+\mathcal O(\frac1n)$, and $n\left\lceil\log_2(n)\right\rceil$ grows faster than that, by a factor $\frac1{\ln(2)}>1$.

REPLY [3 votes]: Approximation of the Probability
The probability that a particular $j$ values have been missed is
$$
N_j=\overbrace{\quad\binom{n}{j}\quad}^{\substack{\text{number of}\\\text{ways to choose}\\\text{the $j$ missed}}}\overbrace{\left(1-\frac jn\right)^m\vphantom{\binom{n}{j}}}^{\substack{\text{probability of}\\\text{missing those}\\\text{$j$ values}}}\tag1
$$
The Principle of Inclusion-Exclusion says that the probability of missing at least one value is
$$
p(m,n)=\sum_{j=1}^n(-1)^{j-1}\binom{n}{j}\left(1-\frac jn\right)^m\tag2
$$
Therefore,
$$
\begin{align}
p(n\log_2(n),n)
&=\sum_{j=1}^n(-1)^{j-1}\binom{n}{j}\left(1-\frac jn\right)^{n\log_2(n)}\tag{3a}\\
&\sim\sum_{j=1}^n(-1)^{j-1}\binom{n}{j}e^{-j\log_2(n)}\tag{3b}\\[3pt]
&=\left(1-\left(1-e^{-\log_2(n)}\right)^n\right)\tag{3c}\\[9pt]
&=\left(1-\left(1-\tfrac1{n^{1/\hspace{-1pt}\log(2)}}\right)^n\right)\tag{3d}\\[9pt]
&\sim n^{1-1/\hspace{-1pt}\log(2)}\tag{3e}\\[12pt]
&\to0\tag{3f}
\end{align}
$$
Explanation:
$\text{(3a)}$: apply $(2)$
$\text{(3b)}$: Poisson approximation
$\text{(3c)}$: Binomial Theorem
$\text{(3d)}$: $e^{-\log_2(n)}=\frac1{n^{1/\hspace{-1pt}\log(2)}}$
$\text{(3e)}$: Binomial approximation
$\text{(3f)}$: $1-1/\log(2)\approx-0.442695$

Comparison of Values
$$
\begin{array}{c|c|c}
n&p(n\log_2(n),n)&n^{1-1/\hspace{-1pt}\log(2)}\\\hline
10&0.275663&0.360832\\
100&0.118840&0.130200\\
1000&0.045682&0.046980\\
10000&0.016798&0.016952
\end{array}
$$