Proof by induction that the sum of transposed matrices equals the transposition of the sum of the matrices Let $A_k \in \mathbb{K}^{p\times r}$ for all $k \in \underline{n}$.
Show that $(A_1 + A_2 + \ldots + A_n)^T = A_1^T + A_2^T + \ldots + A_n^T.$
Or rather $\sum_{k=1}^{n}A_k^T = (\sum_{k=1}^{n}A_k)^T.$
I tried doing that with induction, but I'm having problems in the induction step:
$$\sum_{k=1}^{n+1}A_k^T = \sum_{k=1}^{n}A_k^T + A_{n+1}^T = \left(\sum_{k=1}^{n}A_k\right)^T + A_{n+1}^T$$
How do I continue? How do I get that last term into the sum? I feel like I'd have to use the statement I want to prove to prove it which is nonsense of course.

You're asking about proving
$$\sum_{i=1}^{n}A_i^T = (\sum_{i=1}^{n}A_i)^T \tag{1}\label{eq1A}$$
for all integers $n \ge 1$. With the first base case of $n = 1$, the LHS and RHS are both just $A_1^T$. Next, as basically suggested in Jair Taylor's comment, it's useful to prove the $n = 2$ on its own, i.e.,
$$B^T + C^T = (B + C)^T \tag{2}\label{eq2A}$$
for any $B, C \in \mathbb{K}^{p\times r}$. Similar to what's stated in If $A$ and $B$ are arbitrary $m\times n$ matrices, show that $^t(A+B)= {}^tA+{}^tB$?, this can be shown by checking component wise by using $X_{i,j}$ to mean the element at the $i$'th row and $j$'th column of the matrix $X$. By also using the entrywise sum property of matrix addition, the LHS is $(B^T + C^T)_{i,j} = (B^T)_{i,j} + (C^T)_{i,j} = B_{j,i} + C_{j,i}$, while the RHS is $\left((B + C)^T\right)_{i,j} = (B + C)_{j,i} = B_{j,i} + C_{j,i}$, i.e., they're equal.
Next, for the induction step, assume for some integer $k \ge 2$ that \eqref{eq1A} is true for $n = k$. We then have using this (in the second line below), and \eqref{eq2A} (in the third line below), that
$$\begin{equation}\begin{aligned}
\sum_{i=1}^{k+1}A_i^T & = \sum_{i=1}^{k}A_i^T + A_{k+1}^T \\
& = (\sum_{i=1}^{k}A_i)^T + A_{k+1}^T \\
& = ((\sum_{i=1}^{k}A_i) + A_{k+1})^T \\
& = (\sum_{i=1}^{k+1}A_i)^T
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
i.e., \eqref{eq1A} is also true for $n = k+1$. Thus, by induction, \eqref{eq1A} is true for all integers $n \ge 1$.
