For the convergence, we write
$$ X_{n,i} := \frac{1}{n^i} \sum_{1 \leq k_1 < \cdots < k_i \leq n} A\bigl(\tfrac{k_1}{n}\bigr)\cdots A\bigl(\tfrac{k_i}{n}\bigr). $$
(Here, we set $X_{n,0} := I_d$ and $X_{n,i} := 0$ for $i > n$.) Then we may write the product as
$$ E_n := \biggl(I_d + \frac{A\bigl(\tfrac{1}{n}\bigr)}{n} \biggr) \cdots \biggl(I_d + \frac{A\bigl(\tfrac{n}{n}\bigr)}{n}\biggr)
= \sum_{i=0}^{\infty} X_{n,i}. $$
Since the operator norm of $X_{n,i}$ admits the uniform bound
$$ \| X_{n,i} \| \leq \frac{1}{i!} \|A\|_{\sup}^i $$
with $\|A\|_{\sup} := \sup_{0 \leq t \leq 1} \|A(t)\|$, Weierstrass M-test shows that $E_n$ converges provided $X_{n,i}$ converges as $n\to\infty$ for each $i$. But if we write $\mathcal{T}_i := \{ (t_1, \cdots, t_i) : 0 \leq t_1 \leq \cdots \leq t_i \leq 1\}$, then
$$ \lim_{n\to\infty} X_{n,i} = \int_{\mathcal{T}_i} A(t_1) \cdots A(t_i) \, \mathrm{d}t_1 \cdots \mathrm{d} t_i, $$
the desired conclusion follows. Moreover, we obtain:
$$ \lim_{n\to\infty} \biggl(I_d + \frac{A\bigl(\tfrac{1}{n}\bigr)}{n} \biggr) \cdots \biggl(I_d + \frac{A\bigl(\tfrac{n}{n}\bigr)}{n}\biggr) = \sum_{i=0}^{\infty} \int_{\mathcal{T}_i} A(t_1) \cdots A(t_i) \, \mathrm{d}t_1 \cdots \mathrm{d} t_i. $$
If $\{ A(t) \}_{t \in [0, 1]}$ commutes, then the above integral reduces to
$$ \int_{\mathcal{T}_i} A(t_1) \cdots A(t_i) \, \mathrm{d}t_1 \cdots \mathrm{d} t_i
= \frac{1}{n!} \biggl( \int_{0}^{1} A(t) \, \mathrm{d}t \biggr)^i $$
and thus the limit of $E_n$ becomes the matrix exponential $\exp\bigl( \int_{0}^{1} A(t) \, \mathrm{d}t \bigr)$. But we do not expect this to happen in general.
Since, $$f_n'(x)=nx^{n-1}\ln x +x^{n-1}$$ we get:
$$f_n^{(n)}(x)=n f_{n-1}^{(n-1)}(x)+(n-1)!$$
So if $a_n(x)=f_n^{(n)}(x)$ then:
$$a_n(x)=na_{n-1}(x)+(n-1)!$$
$$a_n(x)=n(n-1)a_{n-2}(x)+n(n-2)!+(n-1)!$$
Repeating:
$$a_n(x)=n!\log x+\sum_{k=1}^{n} \frac{n!}{k}$$
So:
$$\frac{a_n(1/n)}{n!}=-\log (n) +\sum_{k=1}^n\frac1k,$$
which converges to $\gamma,$ the Euler–Mascheroni constant.
Best Answer
Here's a different approach based on the matrix logarithm. For $\|A\|<1$ we have the power series form $\log(I+A)=\sum_{k\geq1}(-1)^{k+1}A^k/k$, so for all sufficiently large $n$ we have $$\log\left((I+X/n)^n\right)=n\log(I+X/n)=n[X/n+o(1/n)]=X+o(1).$$ The desired result follows by the continuity of the matrix exponential.