Triangularization of matrices over algebraically closed field A friend of mine is studying physics in first semester and for his next assignment, he has to prove the following theorem:

Let $V$ be a finite dimensional vector space over an algebraically closed field $K$. Further, let $f: V \to V$ be an endomorphism. Then there exists a basis $B$ of $V$, such that $\mbox{Mat}_{B,B}(f)$ is an upper triangular matrix.

Now this theorem really stumbles me, because I know two proofs of it but they are way beyond first semester. They have only introduced elementary matrix/basis manipulation, basis change theorems and they know theorems about the existence of eigenvectors and eigenvalues ($K$ is algebraically closed so there has to exist an eigenvector). Is there a way to prove this theorem just with the mentioned work tools?
Thanks for your help!
 A: Let $A$ be the matrix of $f$ with respect to a basis. You need to find an invertible matrix $S$ and an upper triangular matrix $T$ such that $A=STS^{-1}$.
Since the base field is algebraically closed, we can find an eigenvalue $\lambda$ and an eigenvector $v$. Complete $v$ to a basis of $V$, say $\{v=v_1,v_2,\dots,v_n\}$, and let $S_0=[v_1\ v_2\ \dots\ v_n]$. Then
$$
S_0^{-1}AS_0=
\begin{bmatrix}
\lambda & \mathbf{x}^T \\
\mathbf{0} & A_1
\end{bmatrix}
$$
for some vector $\mathbf{x}\in K^{n-1}$ and some $(n-1)\times(n-1)$ matrix $A_1$. By induction hypothesis, there is an invertible $(n-1)\times(n-1)$ matrix $S_1$ such that
$$
T_1=S_1^{-1}A_1S_1
$$
is upper triangular. Consider
$$
\hat{S}_1=
\begin{bmatrix}
1 & \mathbf{0}^T \\
\mathbf{0} & S_1
\end{bmatrix}
$$
Then
$$
\hat{S}_1^{-1}=
\begin{bmatrix}
1 & \mathbf{0}^T \\
\mathbf{0} & S_1^{-1}
\end{bmatrix}
$$
and
\begin{align}
\hat{S}_1^{-1}S_0^{-1}AS_0\hat{S}_1&=
\begin{bmatrix}
1 & \mathbf{0}^T \\
\mathbf{0} & S_1^{-1}
\end{bmatrix}
\begin{bmatrix}
\lambda & \mathbf{x}^T \\
\mathbf{0} & A_1
\end{bmatrix}
\begin{bmatrix}
1 & \mathbf{0}^T \\
\mathbf{0} & S_1
\end{bmatrix}\\
&=
\begin{bmatrix}
1 & \mathbf{0}^T \\
\mathbf{0} & S_1^{-1}
\end{bmatrix}
\begin{bmatrix}
\lambda & \mathbf{x}^TS_1\\
\mathbf{0} & A_1S_1
\end{bmatrix}
\\
&=
\begin{bmatrix}
\lambda & \mathbf{x}^TS_1\\
\mathbf{0} & S_1^{-1}A_1S_1
\end{bmatrix}
\\
&=
\begin{bmatrix}
\lambda & \mathbf{x}^TS_1\\
\mathbf{0} & T_1
\end{bmatrix}
\end{align}
is upper triangular.
