which detail?

In theory erase could return a move iterator, meaning that you could omit the call to std::move. This wouldn't be backwards compatible though so not going to happen.

wait, how is this supposed to work?

   a[i] = std::move(a.erase(n-1));

There is no erase that takes an index, so I assume that n = a.end(). Also it is missing a dereference:

   a[i] = std::move(*a.erase(a.end()-1));

but erasing the one-before-the-end returns the (new) end iterator, which obviously is not referenceable. In general, after calling erase, it is too late to access the erased element.

You want something like:

  template<class Container, class Iter>
  auto erase_and_return(Container&& c, Iter pos)
  {
     auto x = std::move(*pos);
     c.erase(pos);
     return x;
  }

Also in the general case it doesn't make sense for erase to return a move iterator.

Thanks for the corrections. I mis-read the documentation and thought erase returned an iterator to the elements erased.