How can I create a Simple page that display the latest node of a content type? I write a little module for loading the content via ajax(one of the used tutorials) and now I need a page for the first visiting(front page). I don't want to do this with Views and looked around, but didn`t found a simple solution for this.

So, I extend the hook_menu () function with:

EDITED

$items['first-node'] = array(
  'page callback' =&gt; 'first_node_callback',
  'access callback' =&gt; TRUE, //EDIT -&gt; 'user_access' creates Menu error
  'type' =&gt; MENU_CALLBACK,
);


and the callback:

function first_node_callback() {
  $query = db_select("node", "n");
  $query-&gt;fields('n', array('nid'))
    -&gt;condition("type", "mytype", "=") 
    -&gt;orderBy("created", "DESC")
    -&gt;range(0, 1); 
  $nids = $query-&gt;execute()-&gt;fetchCol();
  $nodes = node_load_multiple($nids);
  return node_view_multiple($nodes, 'full');
}


EDITED END

Should i load this in the #markup or use a normal return or need i an extra theme output? I'm little bit confused with the possibilities!
Just do the following with node object in your page callback function. No need of returning separate #markup.

return drupal_render(node_view($node));