Why does the base64 of a string contain "\n"?

$ echo -n "apfjxkic-omyuobwd339805ak:60a06cd2ddfad610b9490d359d605407" | base64
YXBmanhraWMtb215dW9id2QzMzk4MDVhazo2MGEwNmNkMmRkZmFkNjEwYjk0OTBkMzU5ZDYwNTQw
Nw==

The output has a return before Nw==. What is the correct way to generate base64 in Linux?

Try:

echo -n "apfjxkic-omyuobwd339805ak:60a06cd2ddfad610b9490d359d605407" | base64 -w 0

From man base64:

-w, --wrap=COLS
Wrap encoded lines after COLS character (default 76). Use 0 to disable line wrapping.

A likely reason for 76 being the default is that Base64 encoding was to provide a way to include binary files in e-mails and Usenet postings which was intended for humans using monitors with 80 characters width. Having a 76-character width as default made that usecase easier.

This is inferior to Kamil's answer on systems which support the -w option to base64, but for cases when that is not available (e.g. Alpine Linux, an Arch Linux initramfs hook, etc.), you can manually process the output of base64:

base64 some_file.txt | tr -d \\n

This is the brute-force approach; instead of getting the program to co-operate, I am using tr to indiscriminately strip every newline on stdout.