Another approach is to use ls when reading the file list within a directory so as to give you what you want, i.e. "just the file name/s". As opposed to reading the full file path and then extracting the "file name" component in the body of the for loop.
Example below that follows your original:
for filename in $(ls /home/user/)
do
echo $filename
done;
If you are running the script in the same directory as the files, then it simply becomes:
for filename in $(ls)
do
echo $filename
done;