Show that discretization matrix is positive-definite 
Given the following coefficient matrix $A^h$, resulting from the finite difference approximation of the biharmonic equation on a mesh with mesh size $h$:
\begin{equation*}
    A^h = \frac{1}{h^4}
    \begin{pmatrix}
    5 & -4 & 1 & & \\
    -4 & 6 & \ddots & \ddots & \\
    1 & \ddots & \ddots & \ddots & 1 \\
     & \ddots & \ddots & 6 & -4 \\
      & & 1 & -4 & 5
    \end{pmatrix}.
\end{equation*}
Show that $A^h$ is symmetric, positive-definite.

Of course it is symmetric, since $(A^h)^T = A^h$. In order to show that it is also positive-definite, we need to let $\mathbf{u}\in \mathbb{R}^n\setminus\{\mathbf{0}\}$ and have that $\mathbf{u}^TA\mathbf{u} >0$. I wrote it out:
\begin{align}
  \mathbf{u}^TA\mathbf{u} &=  \frac{1}{h^4}
  \begin{pmatrix}
     u_1 & \cdots & u_n
  \end{pmatrix}
   \begin{pmatrix}
    5 & -4 & 1 & & \\
    -4 & 6 & \ddots & \ddots & \\
    1 & \ddots & \ddots & \ddots & 1 \\
     & \ddots & \ddots & 6 & -4 \\
      & & 1 & -4 & 5
   \end{pmatrix}
   \begin{pmatrix}
     u_1\\ \\ \vdots\\ \\ u_n
   \end{pmatrix} \\[2ex]
  &= \frac{1}{h^4}
  \begin{pmatrix}
     u_1 & \cdots & u_n
  \end{pmatrix}
  \begin{pmatrix}
    5u_1 - 4u_2 + u_3 \\
    -4u_1 + 6u_2 -4u_3 + u_4 \\
   u_1 - 4u_2 + 6 u_3 - 4u_4 + u5 \\
   \vdots \\
   u_{i-2} - 4u_{i-1} + 6u_i - 4u_{i+1} + u_{i+2} \\
   \vdots 
  \end{pmatrix} \\[2ex]
  &= \frac{1}{h^4}\left[5u_1^2 - 8u_1u_2 + 2u_1u_3 + 6u_2^2 - 8u_2u_3 + 2u_2u_4 + 6u_3^2 -8u_3u_4 + 2 u_3u_5 + \dots\right]
\end{align}
This where I get stuck. I know that I have to show that above can be expressed as the sum of squares, wich means it is positive. I've tried rewriting it using terms like $(u_1 - u_2)^2$ and $(u_1 + u_2 + u_3)^2$ etc, but I cannot see how it all fits together. Help would be greatly appreciated.

Every symmetric, positive-definite matrix has a square root.
In particular, the root may be asked to be symmetric and positive-definite as well, and then it is uniquely determined.
For the given discretisation matrix $A^h$, which is highly structured, (t-)his square root
$$\sqrt{A^h} \;=\; \frac1{h^2}
    \begin{pmatrix}
   2 & -1 & 0 & &\\ -1 & 2 & -1 & &  & \\
    0 & -1 & \ddots & \ddots &\\  & & \ddots & \ddots & -1 & 0\\
     & & & -1 & 2 & -1\\  & & & 0  & -1 & 2
    \end{pmatrix}.$$
is not so far away, let's say by an educated guess which extends
$$\begin{pmatrix}
5 & -4\\ -4 & 5
\end{pmatrix} \;=\;
\begin{pmatrix}
2 & -1\\ -1 & 2
\end{pmatrix}^2\:.$$
And $\sqrt{A^h}$ is positive-definite:
Cf$\,$
https://en.wikipedia.org/wiki/Definite_matrix#Examples  for the $3\times 3$ matrix. The method should generalise to higher dimensions.
And see here on this site:
The full statement including proof of $(7.4.7)$ Theorem starts on page 537 in https://zhilin.math.ncsu.edu/TEACHING/MA580/Stoer_Bulirsch.pdf .

An explicit Sum of squares expression as looked for in the OP is given by
$$(2u_1-u_2)^2 \,+\, \sum_{j=2}^{n-1}\,(-u_{j-1}+2u_j-u_{j+1})^2 \,+\, (2u_n-u_{n-1})^2\tag{SoS}$$
Having the knowledge of the square root $\sqrt{A^h}\,$ it results from expanding
\begin{align}
  h^4\cdot\mathbf{u}^TA^h\,\mathbf{u} & \:=\: 
 \left(h^2\cdot\mathbf{u}^T\sqrt{A^h}\right) \left(h^2\cdot\sqrt{A^h}\,\mathbf{u}\right) \\[2ex]
  & \:=\:
  \big(2u_1-u_2, \dots, -u_{n-1}+2u_n\big)
  \begin{pmatrix}
    2u_1 -u_2\\ -u_1 +2u_2 -u_3\\
   \vdots \\
   -u_{n-2} +2u_{n-1} -u_n \\ -u_{n-1} +2u_n
  \end{pmatrix}
\end{align}
The expression $(\text{SoS})$ gets zero only if $\mathbf{u}= \mathbf{0}$.
This is equivalent to the linear system $\sqrt{A^h}\,\mathbf{u} = \mathbf{0}$ having only the trivial solution.
