How many ways to do n chose k such that the picks are non-decreasing? Given n that represents the maximum value of all the numbers we can pick from such that
1 <= pick <= n

Given k that represents how many numbers we must pick.
I am interested in finding the number of combinations that will have a non-decreasing order.
For example with n = 10, k = 2 if we write it down we can observe:
Pick#1 Pick#2    
10     10 (only possible value)    
9      9 or 10 (same...)
8      8 -> 10    
.      .
.      .
1      1 -> 10 (all values are possible)

And so we find that for k = 2 the number of such combinations will be n(n+1)/2 = 55.
The problem I am facing is that I cannot derive a general formula that would work for any k.
My best attempt:
If n = 10, k = 3:
If Pick#1 = 1, we get (n+1)/2 = 55 combinations possible.
If Pick#1 = 2, we get (n(n+1)/2) - n combinations.
If Pick#1 = 3, we get (n(n+1)/2) - n - (n - 1) combinations.
If Pick#1 = 4, we get (n(n+1)/2) - n - (n - 1) - (n - 2) combinations.
.....
Which can sum to give a total of:
n * (n(n+1)/2) - n(n+1)/2 = (n-1) * (n(n+1)/2) combinations.

I attempt to generalize by recognizing that this pattern will repeat and by writing:
((n-1)^k-2) * (n(n+1)/2) total combinations.

But this is wrong as my test case is that n = 93, k = 8 should equal: 186087894300.
Pick#1  Pick#2 Pick#3    
1       10     10 
        9      9 or 10
        8      8 -> 10    
        .      .
        .      .
        1      1 -> 10

2       10     10 
        9      9 or 10
        8      8 -> 10    
        .      .
        .      .
        2      2 -> 10

3       10     10 
        9      9 or 10
        8      8 -> 10    
        .      .
        .      .
        3      3 -> 10

Thanking you in advance for any help!
 A: You want the number of non-decreasing $k$-tuples of integers chosen from the set $\{1,\ldots,n\}$. Let $\langle a_1,\ldots,a_k\rangle$ be such a $k$-tuple. Let $d_0=a_1-1$ and $d_k=n-a_k$, and for $i=1,\ldots,k-1$ let $d_i=a_{i+1}-a_i$. Note that the $(k+1)$-tuple $\langle d_0,\ldots,d_k\rangle$ is uniquely determined by $\langle a_1,\ldots,a_k\rangle$ and vice versa: from $\langle d_0,\ldots,d_k\rangle$ you can reconstruct $\langle a_1,\ldots,a_k\rangle$. What $(k+1)$-tuples $\langle d_0,\ldots,d_k\rangle$ are possible? Clearly each $d_i$ is a non-negative integer, and
$$\sum_{i=0}^kd_i=(a_1-1)+\sum_{k=1}^{k-1}(a_{i+1}-a_i)+(n-a_k)=n-1\;.$$
You can check that any $(k+1)$-tuple $\langle d_0,\ldots,d_k\rangle$ of non-negative integers summing to $n-1$ comes from a non-decreasing $k$-tuple $\langle a_1,\ldots,a_k\rangle$ whose terms are chosen from $\{1,\ldots,n\}$. Thus, there are exactly as many of your $k$-tuples as there are solutions in non-negative integers to the equation 
$$x_0+x_1+\ldots+x_k=n-1\;.$$
This is a standard stars-and-bars problem whose solution is
$$\binom{(n-1)+(k+1)-1}{(k+1)-1}=\binom{n+k-1}k=\binom{n+k-1}{n-1}\;;$$
the linked article gives a reasonably clear derivation of this formula.
