Average training instances sampled with bagging

The book Hands-On Machine Learning has a section on Out-of-Bag Evaluation related to Decision Trees, where it's stated that,

By default a BaggingClassifier samples m training instances with replacement (bootstrap=True), where m is the size of the training set. This means that only about 63% of the training instances are sampled on average for each predictor.

I am curious how they arrived at 63%. Here's what I have so far,

Let $$X$$ be a random variable representing the fraction of the training set sampled with replacement with size m. Hence, $$X$$ takes values $$\frac{1}{m}, \frac{2}{m}, ..., 1$$ Then, the PMF is as follows,

$$P(X=\frac{i}{m}) = {m\choose i} * \frac{1}{m^m}$$

Then, the average number of training samples with replacement will be,

$$E[X] = \sum_{i=1}^{m} {m\choose i} * \frac{1}{m^m} * \frac{i}{m}$$

Is this the right way to derive the 63%? I am not fully convinced the PMF is correct because the PMF doesn't seem to sum to 1.

$$\sum_{i=1}^{m}P(X=\frac{i}{m}) = (\frac{2}{m})^m - \frac{1}{m^m} = \frac{2^m-1}{m^m} \text{(using Binomial Theorem)}$$

If you randomly draw one instance from a dataset of size $$m$$, each instance in the dataset obviously has probability $$\frac{1}{m}$$ of getting picked, and therefore it has a probability $$1 – \frac{1}{m}$$ of not getting picked.
If you draw $$m$$ instances with replacement, all draws are independent and therefore each instance has a probability $$(1 – \frac{1}{m})^m$$ of not getting picked. Now let's use the fact that $$e^x$$ is equal to the limit of $$(1 + \frac{x}{m})^m$$ as $$m$$ approaches infinity.
So if $$m$$ is large, the ratio of out-of-bag instances will be about $$e^{–1}\approx0.37$$. So roughly $$63\%$$ $$(1 – 0.37)$$ will be sampled.