Let $K \in L_{loc}^{\infty}(\mathbb{R}^+)$, which means that $K$ is bounded on every segment of $\mathbb{R}^+$. For a $n \in \mathbb{N}$ we denote $K^{\star n} = K \star \dots \star K$(n times) where $\star$ denotes the convolution operation. Show that $\forall n \in \mathbb{N}, \forall a >0$ we have: $$|K_n(x)| \leq M(a)^n \frac{x^{n-1}}{(n-1)!} $$ where $M(a) = \underset{x \in [0,a]}{|K(x)|}$.
Now approaching the way to do this seems to be by induction. Starting off for $n=1$, we have $$K\star K(x) = \int_{\mathbb{R}}K(x-y)K(y)dy = \int_{\mathbb{R}^+}K(x-y)K(y)dy = \int_{0}^xK(x-y)K(y)dy$$ $$ \leq \int_{0}^x \underset{z \in [0,x]}{\sup |K(z)|}\underset{z \in [0,x]}{\sup|K(z)|} dy = \underset{z \in [0,x]}{\sup |K(z)|}^2\int_{o}^{x}dy= \underset{z \in [0,x]}{\sup |K(z)|}^2x$$
So the formula seems to be true for the case $n=2$. Now what I struggle to see is how come $|K \star K(x) |\leq M(a)^n \frac{x^{n-1}}{(n-1)!}$ for any $a >0$. I am unable to prove it because honestly this seems to be false to me. Any help will be appreciated.