Doob's Martingale Decomposition -- Proving that the Martingale component is indeed a Martingale

The Martingale-Part of the Doob decomposition for a stochastic process $$(X_n)_n$$ and filtration $$(\mathcal F_n)_n$$ is $$M_n=X_0+\sum_{k=1}^n\bigl(X_k-\mathbb{E}[X_k\,|\,\mathcal{F}_{k-1}]\bigr)$$ (see e.g. here).

I want to prove that this is indeed a Martingale and suceeded by showing $$\mathbb E[M_n-M_{n-1}| \mathcal F_{n-1}] =0$$. However, I failed to show $$\mathbb E[M_n| \mathcal F_{n-1}] =M_{n-1}$$ and wanted to know where my mistake is.

My attempt is: $$\mathbb E[M_n| \mathcal F_{n-1}] = X_0 + \sum_{k=1}^n(\mathbb E[X_k | \mathcal F_{n-1}] - \mathbb E\big[ \mathbb E[X_k | \mathcal F_{k-1}] | \mathcal F_{n-1}\big]) = (*).$$

Now, note that for $$k: $$\mathbb E\big[ \mathbb E[X_k | \mathcal F_{k-1}] | \mathcal F_{n-1}\big]) = \mathbb E[X_k | \mathcal F_{n-1}] = X_k,$$

and for $$k=n$$: $$\mathbb E\big[ \mathbb E[X_k | \mathcal F_{k-1}] | \mathcal F_{n-1}\big]) = \mathbb E[X_n | \mathcal F_{n-1}],$$ so that all terms of the sum in $$(*)$$ are 0. Therefore, we get

$$(*) = X_0.$$

However, we should get $$(*) = M_{n-1}$$.

Where am I making my mistake?

When $$k=n$$, both terms become $$X_n$$, thus it is $$X_n-X_n=0$$. When $$k, it remains same. Therefore, you will have $$M_{n-1}$$.
To prove the martingale property (you also have to show $$(\mathscr{F}_n)_{n \in \mathbb{N}_0}$$-adaptedness and integrability) it is sufficient to notice that $$M_n-M_{n-1}=X_n-E[X_n|\mathscr{F}_{n-1}]$$ so that if you take the conditional expectation wrt $$\mathscr{F}_{n-1}$$ we get $$E[M_n-M_{n-1}|\mathscr{F}_{n-1}]=0$$ because $$E[X_n|\mathscr{F}_{n-1}]$$ is, of course, $$\mathscr{F}_{n-1}$$-measurable.
Your mistake lies into claiming $$E[E[X_k|\mathscr{F}_{k-1}]|\mathscr{F}_{n-1}]\stackrel{(?)}{=}E[X_k|\mathscr{F}_{n-1}]$$. This is not true: note that $$k\leq n$$ so all $$E[X_k|\mathscr{F}_{k-1}]$$ are $$\mathscr{F}_{n-1}$$-measurable. So the equality is in fact $$E[E[X_k|\mathscr{F}_{k-1}]|\mathscr{F}_{n-1}]=E[X_k|\mathscr{F}_{k-1}]$$. Also note that all $$X_k,k\leq n-1$$ are $$\mathscr{F}_{n-1}$$-measurable, but $$X_n$$ is not. So we have \begin{aligned}E[M_n|\mathscr{F}_{n-1}]&=X_0+(E[X_n|\mathscr{F}_{n-1}]-E[E[X_n|\mathscr{F}_{n-1}]|\mathscr{F}_{n-1}])+\sum_{k\leq n-1}(X_k-E[E[X_k|\mathscr{F}_{k-1}]|\mathscr{F}_{n-1}])=\\ &=X_0+0+\sum_{k\leq n-1}(X_k-E[X_k|\mathscr{F}_{k-1}])=\\ &=M_{n-1}\end{aligned}