Conclusion of $\dim X_n = \dim H(X_n)$ It's a part of proof that Euler characteristic of complex chain equal to the Euler characteristic of the homology group.
Let $0\longrightarrow X_{n+1}\overset{f_{n+1}}\longrightarrow X_n\overset{f_n}\longrightarrow X_{n-1}\overset{f_{n-1}}\longrightarrow \cdots\longrightarrow X_1\overset{f_1}\longrightarrow X_0\longrightarrow 0$ denote a chain complex of finite dimensional vector spaces. I wan't to prove that $\dim X_n = \dim H(X_n)$, where $H(X_n)$ is the homology group.
First of all, for a short exact sequence of vector spaces $0\longrightarrow  U\overset{f}\longrightarrow V\overset{g}\longrightarrow W \longrightarrow 0$, we have $\dim V = \dim U + \dim W$ (already proved). Then, for
$$\cdots \longrightarrow X_{k+1}\overset{f_{k+1}}\longrightarrow X_k\overset{f_k}\longrightarrow X_{k-1}\overset{f_{k-1}}\longrightarrow \cdots$$
we take the s.e.s.
$$
0\longrightarrow Im(f_{k+1})\longrightarrow Ker(f_k)\longrightarrow H(X_k)=\frac{Ker(f_k)}{Im(f_{k+1})}\longrightarrow 0
$$
then we have
$$\dim H(X_k)= \dim \ker f_k - \dim \mathrm{Im} f_{k+1}\:\:\:\:\:(*)$$
From the s.e.s. $0\longrightarrow \ker(f_k)\longrightarrow X_k\longrightarrow \mathrm{Im}(f_k)\to 0$, we have
$$\dim X_k = \dim \ker f_k + \dim \mathrm{Im} f_k \:\:\:\:\:(**)$$
I omitted the boring details. I'm stuck on the following :
Combining $(**)$ with $(*)$, we have
$$
\dim(X_k)+ \dim \mathrm{Im} f_k= \dim H(X_k) + \dim \mathrm{Im}f_{k+1}.
$$
I think I'm missing something simple to conclude that $\dim X_k = \dim H(X_k)$ and finish the question, someone can help me?

What if we define the following chain complex $0\to k^n\xrightarrow{\mathrm{id}} k^n\to 0$? This chain complex has no homology, but the vector spaces are nonzero.
