Print a file, skipping the first X lines, in Bash

You'll need tail. Some examples:

$ tail great-big-file.log
< Last 10 lines of great-big-file.log >

If you really need to SKIP a particular number of "first" lines, use

$ tail -n +<N+1> <filename>
< filename, excluding first N lines. >

That is, if you want to skip N lines, you start printing line N+1. Example:

$ tail -n +11 /tmp/myfile
< /tmp/myfile, starting at line 11, or skipping the first 10 lines. >

If you want to just see the last so many lines, omit the "+":

$ tail -n <N> <filename>
< last N lines of file. >

Easiest way I found to remove the first ten lines of a file:

$ sed 1,10d file.txt

In the general case where X is the number of initial lines to delete, credit to commenters and editors for this:

$ sed 1,Xd file.txt

If you want to skip first two line:

tail -n +3 <filename>

If you want to skip first x line:

tail -n +$((x+1)) <filename>

If you have GNU tail available on your system, you can do the following:

tail -n +1000001 huge-file.log

It's the + character that does what you want. To quote from the man page:

If the first character of K (the number of bytes or lines) is a `+', print beginning with the Kth item from the start of each file.

Thus, as noted in the comment, putting +1000001 starts printing with the first item after the first 1,000,000 lines.