Why does this select always run the default case when the first case actually is executed?

I'm trying to get a better understanding of golang channels. While reading this article I'm toying with non-blocking sends and have come up with the following code:

package main
import (
    "fmt"
    "time"
)

func main() {
    stuff := make(chan int)
    go func(){
        for i := 0; i < 5; i ++{
            select {
            case stuff <- i:
                fmt.Printf("Sent %v\n", i)
            default:
                fmt.Printf("Default on %v\n", i)
            }
        }
        println("Closing")
        close(stuff)
    }()
    time.Sleep(time.Second)
    fmt.Println(<-stuff)
    fmt.Println(<-stuff)
    fmt.Println(<-stuff)
    fmt.Println(<-stuff)
    fmt.Println(<-stuff)
}

This will print:

Default on 0
Default on 1
Default on 2
Default on 3
Default on 4
Closing
0
0
0
0
0

While I do understand that only 0s will get printed I do not really understand why the first send does still trigger the default branch of the select?

What is the logic behind the behavior of a select in this case?

Example at the Go Playground