First yes this question is related to the problem I posted before here. But I want to try proving this via induction this time.
Let $H_k(n)$ be the number of vectors $x_1,\dots,x_k$ for which each $x_i$ is a positive integer satisfying $1\leq x_i\leq n$ and $x_1\leq x_2 \leq \dots \leq x_k$. Prove \begin{equation*} \begin{aligned} H_1(n) & = n & \\ \\ H_k(n) & = \sum\limits_{j=1}^n H_{k-1}(j) & k>1 \end{aligned} \end{equation*}
So the proof is the same for $i=1,2$ which I will provide below. From the previous post, Let $(x_1,\dots,x_k)$ be a vector such that $x_i\geq 0$ and $x_i\in \mathbb{Z}$ and $1\leq x_i\leq n$.
When $i=1$, then the vector is of length 1. Since $x_1$ is an integer between (and including) 1 and $n$, there are $n$ different choices of integers. Hence $H_1(n)=n$.
When $i=2$, then the vector is of length 2. Since $x_1$ is an integer between (and including) 1 and $n$, we have $n$ options. Now for $x_2$. We know $x_1\leq x_2$. This implies whatever $x_1$ equals $x_2$ has $n-x_1+1$ options (because $x_1$ can be the same as $x_2$). Hence $$ H_2(n) = \sum\limits_{x_1\in\{1,2,\dots,n\}} (n-x_1+1) = 1+2+\dots+n = \dfrac{n(n+1)}{2} $$
Now here is the different part. Assume that the sum is true for $i=k$. i.e. $$ H_k(n) = \sum\limits_{j=1}^n H_{k-1}(j)$$ We want to show that $$ H_{k+1}(n) = \sum\limits_{j=1}^n H_{k}(j)$$ So here is what I got so far $$ \sum\limits_{j=1}^n H_{k}(j) = \sum\limits_{j=1}^n \sum\limits_{\ell=1}^j H_{k-1}(\ell)$$ I know I can repeat this continuously until I get $H_1(\alpha)$ where $\alpha$ is some positive integer. But that doesn't get me anywhere.
I have also tried doing the following: \begin{equation*} \begin{aligned} \sum\limits_{j=1}^n H_{k}(j) & = H_k(1) + H_k(2) + H_k(3) +\dots+H_k(n) \\ & = \sum\limits_{j=1}^1 H_{k-1}(j) + \sum\limits_{j=1}^2 H_{k-1}(j) + \sum\limits_{j=1}^3 H_k(j) + \dots + \sum\limits_{j=1}^n H_{k-1}(n) \\ & = nH_{k-1}(1) + (n-1)H_{k-1}(2) + (n-2)H_{k-1}(3) + \dots +H_{k-1}(n) \end{aligned} \end{equation*} Again I can repeat this step again with the equation for $H_{k-1}(n)$ but I feel like I am going on a never ending battle. Any suggestions.
I would have done induction on $n$: First, it's easy to see that $H_k(1)=1$ for every $k$, so that $H_k(1)=\sum_{j=1}^1H_{k-1}(1)$, for $k>1$.
Let $\mathcal H_k(n)$ be the set as of tuples $(x_1,\cdots,x_k)$ as in the question. You want to break $\mathcal H_k(n)$ into two subsets of which you can apply induction. In this case, you break the set into $x_k<n$ and $x_k=n$, because then you can reduce $n$, because if $x_k=n$, then you're only $k-1$ variables $x_1,\cdots,x_{k-1}$, so the number of tuples $(x_1,\cdots,x_{k-1},n)$ is just $H_{k-1}(n)$. Next, if $x_k<n$, then all other variables must be $<n$ and so we're counting $H_k(n-1)$ and you can apply induction and conclude that $$H_k(n)=H_{k-1}(n)+H_k(n-1)=\sum_{j=1}^nH_{j-1}(n)$$