I am going through the proof of the existence of strong solution of SDE on Brownian motion and Stochastic Calculus by Karatzas and Shreve. However, there is a single line I really cannot see how to conclude. The original theorem is the following: (It is abbreviated)
Theorem 2.9: Suppose that the coefficients $b(t,x), \sigma(t,x)$ satisfy the global lipschitz and linear growth conditions: \begin{align} &||b(t,x)-b(t,y)||+||\sigma(t,x)-\sigma(t,y)||\leq K||x-y||\\ &||b(t,x)||^2+||\sigma(t,x)||^2\leq K^2(1+||x||^2) \end{align} Then the strong solution of the Stochastic Differential Equation exists. The SDE is $dX_t = b(t,X_t)dt+\sigma(t,X_t)dW_t$.
The proof applies the iteration of approximation. It defines \begin{align*} X_t^{(k+1)} = \xi+ \int_0^tb(s,X_t^{(k)}dt + \int_0^t \sigma(s,X_t^{(k)})dW_t \end{align*}
The question is the following: After some lines it concludes for some constant $L$. \begin{align} E[\max_{0\leq s\leq t}||X_s^{(k+1)}-X_s^{(k)}||^2]\leq L\int_0^t E||X_s^{(k)}-X_s^{(k-1)}||^2 ds \end{align}
Then it says, we iterate this inequality to conclude \begin{align} E[\max_{0\leq s\leq t}||X_s^{(k+1)}-X_s^{(k)}||^2]\leq \max_{[0,T]}E||X_t^{(1)}-\xi||^2\cdot \frac{(Lt)^k}{k!} \end{align} Why there is a $k!$? It seems keep iterating we only get $\max_{[0,T]}E||X_t^{(1)}-\xi||^2\cdot (Lt)^k$.
Thanks in advance