How can I show that $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$?

The matrix $$N=\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}$$ is nilpotent with index 2 of nilpotency: $N^2=0$ so by the binomial formula we have

$$\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}^n=(I_2+N)^n=\sum_{k=0}^n {n\choose k}N^k={n\choose 0}I_2+{n\choose 1}N=I_2+nN=\begin{pmatrix} 1& n\\ 0 & 1 \end{pmatrix}$$

Use induction on $n$.

(1) Prove the base case (trivial), perhaps even establish the case for $n = 2$ (two base cases here are not necessary, but as you found, it helps reveal the pattern.)

(2) Then assume it holds for $n = k$.

(3) Finally, show that from this assumption, it holds for $n = k+1$.

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You've established the base case(s). Now, (2) assume the inductive hypothesis (IH) $$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^k = \begin{pmatrix} 1 & k \\ 0 & 1\end{pmatrix}.$$

Then, $$\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}^{k + 1} = \begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^k \quad \overset{IH}{=} \quad \begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix} \begin{pmatrix} 1 & k \\ 0 & 1\end{pmatrix}=\quad\cdots$$

I think you can take it from here!

Geometrically, your matrix represents a shear transformation that preserves the horizontal direction and shifts the vertical direction by the horizontal direction. What happens geometrically if you apply the same shear transformation $n$ times?