Show that $\sum_{k=1}^n\frac{1}{k}(M_k-M_{k-1})$ is a martingale

Question

This is a problem from UMD probability quals here. I'm stuck in showing the conditional expectation requirement of a martingale. The problem is

Let $(\Omega,F,\{F_n\}_{n\in\mathbb{N}},P)$ be a filtered space and let $\{M_n\}_{n\in\mathbb{N}}$be a $\{F_n\}_{n\in\mathbb{N}}$-martingale which is bounded in $L^\infty(\Omega,F,P)$. For every $n\in\mathbb{N}$, we define $$X_n := \sum_{k=1}^n\frac{1}{k}(M_k-M_{k-1})$$

1. Show that $\{X_n\}_{n\in\mathbb{N}}$ is a $\{F_n\}_{n\in\mathbb{N}}$-martingale.
2. Prove that $\{X_n\}_{n\in\mathbb{N}}$ converges $P$-almost surely and in $L^2(\Omega,F,P)$

Solution: Part 2 is answered in a post here. To answer part 1, we need to check the defintion of a martingale, i.e. check the following three requirements:

1. $X_n$ is $F_n$-measurable for every $n$. This is true because (i) $M_k$ is $F_k$ measurable and $F_k\subset F_n$ for all $k\leq n$ since $F$ is a filtration and so (ii) $X_n$ is a linear combination of $F_n$ measurable terms.
2. $\mathbb{E}|X_n|<\infty$ for all $n$. This is also not hard to show: Apply triangle inequality twice, and use the fact that $\mathbb{E}|M_k|<\infty$ for all $k$ since $M_k$ are martingales. $$\begin{eqnarray} \mathbb{E}|X_n| &=& \mathbb{E}|\sum_{k=1}^n\frac{1}{k}(M_k-M_{k-1})|\\ &\leq & \sum_{k=1}^n\frac{1}{k}\mathbb{E}|M_k-M_{k-1}| \\ &\leq & \sum_{k=1}^n\frac{1}{k}(\mathbb{E}|M_k|+\mathbb{E}|M_{k-1}|)<\infty \end{eqnarray}$$
3. $\mathbb{E}(X_{n+1}|F_n)= X_n$. This is where I'm stuck.