Because the Promise will run the "then" statement when it done its task. And if you use the loop like that, the Promise for i=4 will immediately execute after i=5 executed. This will make the output just only wait for the first one.

The solution for this is run the Promise (i=4) after the Promise (i=5) done using recursive function.

function wait(milliseconds) {
    return new Promise(resolve => setTimeout(resolve, milliseconds))
}

function countDown(count) {
    if (count === 0) return;
    wait(1000)
        .then(() => {
            console.log(count)
            countDown(count - 1)
        })
        .catch((error) => console.log(error));
}

countDown(5);

Execution doesn't stop when you call then on a Promise like it does when you use await. On the first loop a Promise is created whose then method will execute in one second, but the function continues looping because execution hasn't stopped. At the end of the loop you have five Promise objects that have all been created consecutively with no pauses in between, so their one-second timeouts all execute very close to one another.

Here's the await method with a comment calling out where execution stops.

function wait(milliseconds) {
    return new Promise(resolve => setTimeout(resolve, milliseconds));
}

async function countDown(count) {
    for (let i = count; i > 0; i--) {
        await wait(1000); // Execution stops here for one second before continuing
        console.log(i);
    }
}

countDown(5);

Here's something closer to Phúc Đỗ Vương's answer with a comment calling out where execution doesn't stop.

function wait(milliseconds) {
    return new Promise(resolve => setTimeout(resolve, milliseconds));
}

function countDown(count) {
    if (count === 0) {
        return;
    }
    
    wait(1000).then(() => {
        // This code, including the next call to countDown, will execute in one second
        console.log(count);
        countDown(count - 1);
    });

    // The current call to countDown finishes before the next call begins
}

countDown(5);