The defending Super Bowl champions New England Patriots are unstoppable this season and the biggest factor to that success is franchise quarterback Tom Brady, who was named the American Football Conference (AFC) Offensive Player of the Week. 

New England is one of four teams in the National Football League (NFL) that has yet to suffer defeat and is sitting comfortably atop the AFC East at 7–0. 

It is already the 25th time in his career that Brady received the recognition, two more and he will tie the lead which is currently held by Denver Broncos quarterback Peyton Manning, who was named AFC Offensive Player of the Week for 27 times already, ESPN reported. 

Brady is having another career year following their Super Bowl victory last season over the Seattle Seahawks. 

In seven games played, the 38-year-old has 199 completions out of 289 attempts, for 2,410 yards and 20 touchdowns, while only throwing one interception.

The latest victim by Brady and the Pats were the Miami Dolphins whom they defeated a week ago for a 36–7 Week 8 victory. 

Before the loss, the Dolphins were thriving riding a two-game winning streak under interim head coach Dan Campbell. 

But even they could not stop the defending champions who are powered by one of the greatest quarterbacks in the league. 

Brady shined once again in their win over the Dolphins as he completed 26-of-38 passes for 356 yards and four touchdowns, although he was sacked twice. 

The first touchdown pass of Brady against Miami was in the first quarter when he connected with Patriots star tight end Rob Gronkowski for a 47-yard score. 

Brady's next touchdown passes were to running back Dion Lewis in the second quarter (16 yards) and wide receiver Julian Edelman to whom he connected twice in the fourth quarter (10 yards and 16 yards). 

The Pats' next assignment will be a home stand this Sunday against the Washington Redskins who are coming off a 31–30 win over the Tampa Bay Buccaneers. 

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