Select deepest child in jQuery

I don't think you can do it directly but you can try

var s = "#SearchHere";
while($(s + " >div ").size() > 0)
    s += " > div";
alert( $(s).attr('id') );

Here's a slight improvement on the answer from @user113716, this version handles the case when there are no children and returns the target itself.

 (function($) {

    $.fn.deepestChild = function() {
        if ($(this).children().length==0)
            return $(this);

        var $target = $(this).children(),
        $next = $target;

        while( $next.length ) {
          $target = $next;
          $next = $next.children();
        }

        return $target;
    };

}(jQuery));

EDIT: This is likely a better approach than my original answer:

Example: http://jsfiddle.net/patrick_dw/xN6d5/5/

var $target = $('#SearchHere').children(),
    $next = $target;

while( $next.length ) {
  $target = $next;
  $next = $next.children();
}

alert( $target.attr('id') );

or this which is even a little shorter:

Example: http://jsfiddle.net/patrick_dw/xN6d5/6/

var $target = $('#SearchHere').children();

while( $target.length ) {
  $target = $target.children();
}

alert( $target.end().attr('id') ); // You need .end() to get to the last matched set

---

Original answer:

This would seem to work:

Example: http://jsfiddle.net/xN6d5/4/

var levels = 0;
var deepest;

$('#SearchHere').find('*').each(function() {
    if( !this.firstChild || this.firstChild.nodeType !== 1  ) {
        var levelsFromThis = $(this).parentsUntil('#SearchHere').length;
        if(levelsFromThis > levels) {
            levels = levelsFromThis;
            deepest = this;
        }
    }
});

alert( deepest.id );

If you know that the deepest will be a certain tag (or something else), you could speed it up by replacing .find('*') with .find('div') for example.

EDIT: Updated to only check the length if the current element does not have a firstChild or if it does, that the firstChild is not a type 1 node.