LINQ to XML - what am I doing wrong

This is my XML file:

<?xml version="1.0" encoding="utf-8" ?>
<Profiles xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://schemas.mycompany.com/MyApp/2010/03/MySchema">
  <Profile id="0" name="MyProfileName1">
    <Entity name="EntityName1" type="EntityTypeName1">
      <Shared>
        <Property name="SharedPropertyName1" type="SharedPropertyType1" />
        <Property name="SharedPropertyName2" type="SharedPropertyType2" />
      </Shared>
      <PropertyGroup name="PropertyGroupName1">
        <Property name="PropertyName1" type="PropertyType1" />
      </PropertyGroup>
      <PropertyGroup name="PropertyGroupName2">
        <Property name="PropertyName2" type="PropertyType2" />
      </PropertyGroup>
    </Entity>
  </Profile>
  <Profile id="1" name="MyProfileName2">
    <Entity name="EntityName2" type="EntityTypeName2">
       <PropertyGroup name="PropertyGroupName3">
        <Property name="PropertyName3" type="PropertyType3" />
      </PropertyGroup>
    </Entity>
  </Profile>
</Profiles>

I want to return all the "Profile" elements, and all the examples I have
seen show one of two syntax:

static void Main(string[] args)
        {
            Program program = new Program();
            IEnumerable<XElement> profiles =
program.GetProfiles(XDocument.Load("Profile.xml"));

            Console.WriteLine(profiles.Count());
            Console.ReadKey();

            profiles = program.GetProfiles2(XDocument.Load("Profile.xml"));

            Console.WriteLine(profiles.Count());
            Console.ReadKey();
        }

        private IEnumerable<XElement> GetProfiles(XDocument document)
        {
            return document.Descendants("Profile");
        }

        private IEnumerable<XElement> GetProfiles2(XDocument document)
        {
            return document.Element("Profiles").Elements("Profile");
        }

The first method return 0; the second method throws a NullReferenceException
when trying to get the Profiles element. This really shouldn't be this
difficult.