How can I show just the most recent post on my home page with jekyll?

It appears you can also just access the latest post via the first index of site.posts as in:

{%- assign latest_post = site.posts[0] -%}

Latest post: <a href="{{ latest_post.url }}">{{ latest_post.title }}</a>

While site.posts.first works too as mentioned by someone else, the above example also provides a consistent manner for accessing other indices besides just the first (not that you would ever need to). Also, I didn't have enough reputation to add this answer as a comment instead :)

Rather than create a loop, just assign the variable and move on...

{% assign post = site.posts.first %}

(Edit 2018) Since someone wanted to know how to iterate other posts after you've done this:

{% for post in site.posts offset:1 %}
  ... Show the next posts ...
{% endfor %}

If you got here for the question as stated in the title, "How can I show just the most recent post on my home page with jekyll?" and not "how do I show only the latest post in my template," the following might be helpful.

Given a brand new Jekyll version 3.7.3 install with the default minima theme, create a file, _layouts/home.html with the following content:

---
layout: none
---
{{ site.posts.first }}

Causes Jekyll 3.7.3 to show the first post, using the post template, as the home page.

This can be accomplished through the use of limit:

{% for post in site.posts limit:1 %}
... Show the post ...
{% endfor %}

You can also use limit and offset together to "feature" your most recent post:

<h1>Latest Post</h1>
{% for post in site.posts limit:1 %}
... Show the first post all big ...
{% endfor %}
<h1>Recent Posts</h1>
{% for post in site.posts offset:1 limit:2 %}
... Show the next two posts ...
{% endfor %}