How to basic display the UUID of a device in PhoneGap?

Question

phonegap

Comrades, help!
I created an application by default: phonegap create my1
I'm trying to build for Android 2.3.3 (API 10)
In the config.xml I wrote: "android-minSdkVersion" value="10"
On the virtual machine: phonegap logo (robot) and the inscription "Device Ready" "shimmers green (everything seems to be ok.)
Task: you need to display the UUID of the device through the usual alert () after the application starts.
Nothing comes out: no reaction. As soon as I have not tried ....
1. Do I need to put extra for this. Are plugins or CORE plugins already enabled when creating the project?
2. How to start your script STRICTLY AFTER! initialization of all application events, when it will be possible to interact with the API, similar to $(document ).ready() ?
(only without jquery!
Versions:

=====[ PhoneGap ]=========
5.0.0-0.28.1
=====[ Cordova ]=========
5.1.1

In advance, Thanks everyone for your help!

Answer 1

This listener monitors the start of the application:

document.addEventListener( "deviceready", function() {
  alert( device.uuid );
}, false );

well, or display the value in some DOM element

var div = document.getElementById( 'result' );
div.innerHTML = device.uuid;