N-dimensional pyramid numbers Given two inputs, a number n and a dimension d, generate the nth d-dimensional pyramid number.
That was confusing, let me try again.
For d = 1, the numbers start 1,2,3,4,5 and is the number of points in a line n points long.
For d = 2, the numbers start 1,3,6,10,15 and is the number of points in a triangle with side length n, also known as the triangle numbers e.g.
     0
    0 0
   0 0 0
  0 0 0 0

For d=3, the numbers start 1,4,10,20,35 and is the number of points in a pyramid of side n. For d=4, it's a 4-d pyramid, and so on.
Beyond this, visualization gets a bit tricky so you will have to use the fact that the nth d-dimensional pyramid number is equal to the sum of the first n d-1-dimensional pyramid numbers.
For example, the number of dots in a 3-d pyramid of side 5 is the sum of the first 5 triangle numbers: 1+3+6+10+15 = 35.
You can expect reasonable input (within your languages boundaries), although Standard loopholes apply. No builtins explicitly for this purpose (looking at you, Mathematica)
Numbers are 1-indexed, unless you specify otherwise.
Example recursive code in Javascript:
function pyramid(dim,num){                          //declare function
  if(dim == 0){                                     //any 0-dimensional is a single point, so return 1
    return 1;
  } else {                                          //otherwise
    function numbersUpTo(x){                        //helper function to get the numbers up to x
      if(x==0){                                     //no numbers up to 0
        return [];
      } else {                                      //otherwise recurse
        return [x].concat(numbersUpTo(x-1));
      }
    }
    var upto = numbersUpTo(num).map(function(each){ //for each number up to num
      return pyramid(dim-1,each);                   //replace in array with pyramid(dimension - 1,the number)
    });
    return upto.reduce((a,b)=&gt;a+b);                 //get sum of array
  }
}

This is code-golf, so fewest bytes wins.
JavaScript (ES6), 24 bytes
Expects (n)(d).

n=&gt;g=d=&gt;d?n++/d*g(d-1):1

Try it online!