What is the formal analysis with Simple Uniform Hashing that the load factor is $\alpha = \frac{n}{m}$
The mean chain length is the sum of the chain lengths divided by the number of chains. The sum of the chain lengths is, by definition $n$ and the number of chains is $m$. This is by far the easiest way to calculate it, and it's true even if the SUHA doesn't hold.

But since you asked, let's look at the distribution of the chain lengths.

Suppose you're putting $n$ pigeons into $m$ pigeonholes at random. Examine one pigeon hole (say, pigeonhole number 1). The number of pigeons that are assigned to this pigeonhole follows a binomial distribution; the probability that there are $k$ pigeons in that particular pigeonhole is ${n \choose k} p^k (1-p)^{n-k}$ where $p = \frac{1}{m}$.

So the expected number of pigeons in that pigeonhole is:

$$\sum_{k=0}^{n} k {n \choose k} p^k (1-p)^{n-k}$$
$$= \sum_{k=0}^{n} np {{n-1} \choose {k-1}} p^{k-1} (1-p)^{n-k}$$
$$= np \sum_{k=0}^{n-1} {{n-1} \choose {k}} p^k (1-p)^{n-1-k}$$
$$= np \left(p + (1-p)\right)^{n-1}$$
$$= \frac{n}{m}$$

By the way, if we want to reason about the number of chains of length $k$ in a hash table, for large enough $m$, it's well-approximated by a Poisson distribution.

The expected proportion of hash slots with chain length $k$ is:

$$\hbox{Pr}(\lambda,k) \approx \frac{e^{-\lambda} \lambda^k}{k!}$$

where $\lambda$ is the load factor. Note that $\lambda$ is also the mean of a Poisson distribution, so once again, the expected value of the chain length is the load factor.

You can learn a lot from this. For example:


If the load factor is 1, the expected proportion of hash chains of length 0 and the expected proportion of hash chains of length 1 are both $\frac{1}{e}$.
If you have a hash table with $m$ slots, there is a 50% chance of a collision after you insert $\sqrt{m}$ items. (Exercise: prove this!)