91Ó°ÊÓ

When \(n=1\), the given integral becomes: $$ V_1 = \int_{0}^{1} dt_1 = \left[t_1\right]_0^1 = 1-0 = 1, $$ and by the definition, \(\frac{1}{1!} = 1\). Therefore, the relationship holds true for \(n=1\).

Now, we'll assume the relationship holds for \(n=k\): $$ V_k = \int_{0}^{1} \int_{0}^{t_1} \cdots \int_{0}^{t_{k-1}} dt_k \cdots dt_2 dt_1 = \frac{1}{k!}. $$

Our goal is to prove the relationship holds for \(n=k+1\). The integral for \(V_{k+1}\) is given as follows: $$ V_{k+1} = \int_{0}^{1} \int_{0}^{t_1} \cdots \int_{0}^{t_{k}} dt_{k+1} \cdots dt_2 dt_1. $$ Now, note that the integration related to \(t_{k+1}\) is inside the integral for \(V_k\): $$ V_k = \int_{0}^{1} dt_1 \int_{0}^{t_1} dt_2 \cdots \int_{0}^{t_{k-1}} dt_k \int_{0}^{t_k} dt_{k+1}. $$ Thus, one can replace the term for \(V_k\) in the proof of \(V_{k+1}\): $$ V_{k+1} = \int_{0}^{1} \int_{0}^{t_1} \cdots \int_{0}^{t_{k}} \frac{1}{k!} dt_k \cdots dt_2 dt_1. $$ By dividing both sides with the constant \(\frac{1}{k!}\), we get: $$ k! V_{k+1} = \int_{0}^{1} \int_{0}^{t_1} \cdots \int_{0}^{t_{k}} dt_k \cdots dt_2 dt_1. $$ Notice that the resulting iterated integral on the right-hand side corresponds to the \((k+1)\)-dimensional volume \(V_{k+1}\) itself. We can now integrate with respect to \(t_k\): $$ k!\int_{0}^{1} \cdots \int_{0}^{t_2} dt_2 dt_1 \left(\int_{0}^{t_{k}} dt_k\right) = \int_{0}^{1} \cdots \int_{0}^{t_{2}}(t_k) \bigg|_0^{t_{k-1}}\,dt_2 dt_1. $$ Hence, $$ (k! V_{k+1}) = \int_{0}^{1} \cdots \int_{0}^{t_{2}} t_{k-1} dt_2 dt_1. $$ Now, we can proceed with the integration w.r.t. \(t_{k-1}\): $$ (k-1)! (k! V_{k+1}) = \int_{0}^{1} \cdots \int_{0}^{t_{2}} (t_{k-2}) dt_2 dt_1. $$ Proceeding with the integration w.r.t. the remaining variables \(t_1, \dots, t_{k-2}\), we finally arrive at: $$ 1! (2! \cdots (k! V_{k+1})) = \int_{0}^{1} t_1 dt_1 = \frac{1}{2}. $$ From this, we deduce that $$ V_{k+1} = \frac{1}{(k+1)!}. $$ With this, we have shown that the relationship holds for \(n=k+1\). Therefore, by Principle of Mathematical Induction, the relationship holds for all \(n \in \mathbb{N}\).