Neil Strickland's answer is very nice. Let me add a slightly different answer, which might be useful in a similar situations, when a homotopy can't be given as easily.

Notice that the homotopy between $f$ and the identity implies that the map of $f: (\Delta_n,\partial\Delta_n)\to(\Delta_n,\partial\Delta_n)$ is of degree $1$, which then implies surjectivity. It is possible to show that $f$ is of degree $1$, whitout using the homotopy.

We proceed by induction, with $n=0$ being trivial. For a $(n-1)$-dimensional facet $\sigma$ in $\partial\Delta_n$, the map $f_{|\sigma}$ is of degree $1$ by induction. Hence the map $f_{|\partial\Delta_n}$ is also of degree $1$, i.e.
$$(f_{|\partial\Delta_n})_*: H_{n-1}(\partial\Delta_n)\to H_{n-1}(\partial\Delta_n)$$ is the identity. A look at the long exact homology sequence $$\require{AMScd}
\begin{CD}
H_n(\Delta_n)@>{}>> H_n(\Delta_n,\partial\Delta_n) @>{\cong}>> H_{n-1}(\partial\Delta_n)@>{}>>H_{n-1}(\Delta_n)\\
@| @VV{f_*}V @V{\cong}V{(f_{|\partial\Delta_n})_*}V@|\\
0 @. H_n(\Delta_n,\partial\Delta_n)@>{\cong}>>H_{n-1}(\partial\Delta_n)@.0
\end{CD}$$
gives that $f$ is also of degree $1$ as claimed.