fix: use regex.source for pattern to handle regex flags correctly

[peekg]

Description

Fixes regex pattern extraction to properly handle regex flags by using the .source property instead of manually slicing the string representation.

Problem

The previous implementation used .toString().slice(1, -1) to extract the pattern from a regex, which incorrectly included flags in the output when they were present (e.g., /^A$/u would produce ^A$/ instead of ^A$).

Solution

Use the standard .source property which returns only the pattern text without delimiters or flags.

Changes

• Replace pattern.args.regex.toString().slice(1, -1) with pattern.args.regex.source
• Add test case for regex with flags to prevent regression

Testing

Added test case verifying that joi.string().regex(/^A$/u) correctly produces pattern: '^A$' in the swagger output.