Following a record-setting season in 2025, Myles Garrett has been voted the best player in the NFL. The Los Angeles Rams superstar landed at No. 1 on the NFL's Top 100 Players of 2026, beating out Josh Allen for the top spot.
It's the highest Garrett has ever ranked on the annual list; his previous high was eighth (2025).
As a member of the Browns last season, Garrett recorded 23 sacks, setting the NFL's single-season record. He also led the league with 33 tackles for a loss, earning his second Defensive Player of the Year award and fifth AP first-team All-Pro selection.
No. 1 on the NFL Top 100 Players of 2026…@RamsNFL DE Myles Garrett! @NFLFilms pic.twitter.com/JBupkPJHbv
— NFL (@NFL) September 3, 2026
Garrett was one of three Rams to land in the top 10 of this year's list,
joining Matthew Stafford (No. 4) and Puka Nacua (No. 7).
Here's the full top 10 for 2026.
Players 1-10 on the NFL Top 100 👏 pic.twitter.com/4ZB6NE5jqO
— NFL (@NFL) September 3, 2026
The last time a defensive player was voted No. 1 on the Top 100 Players list was Aaron Donald in 2029, with J.J. Watt also accomplishing the feat in 2015. Garrett and Donald, of course, will be teammates this season after Donald came out of retirement.
The Rams acquired Garrett in a trade with the Browns this offseason, sending Cleveland Jared Verse, as well as a first-, second- and third-round pick. As his peers said in the video above, he's a dominant pass rusher who's borderline unblockable, earning the unquestioned respect of players around the league – whether it's opposing linemen or fellow defenders.
This article originally appeared on Rams Wire: Myles Garrett lands at No. 1 on NFL's Top 100 Players of 2026
