Looking at the exponential series, we have $\frac{r^k}{k!}\leq e^r$. Hence we have $$ \frac{r^{m-k-r}}{(m-2k-r)!r!k!} = \frac{1}{r!}\frac{r^k}{k!}\frac{r^{m-2k-r}}{(m-2k-r)!} \leq\frac{e^{2r}}{r!}=\mathcal{O}(1). $$ From this we obtain $$ \#End(T_n)\leq Cn! n^3\max_m\frac{m^{n-m}}{(n-m)!} \leq Cn! n^3\max_m\frac{m^{n-m}}{(n-m)!}. $$ Put $m=\alpha n$. Then $$ \frac{m^{n-m}}{(n-m)!}\leq \left(\frac{e\alpha}{1-\alpha}\right)^{(1-\alpha)n}. $$ The maximum of $\left(\frac{e\alpha}{1-\alpha}\right)^{(1-\alpha)}$ is about $1.783$, thus for large $n$ we have $\#End(T_n)\leq 1.79^n n!$, which is a lot smaller than $n^n\sim e^n n!\sqrt{2\pi n}$.

Looking at the exponential series, we have $\frac{r^k}{k!}\leq e^r$. Hence we have $$ \frac{r^{m-k-r}}{(m-2k-r)!r!k!} = \frac{1}{r!}\frac{r^k}{k!}\frac{r^{m-2k-r}}{(m-2k-r)!} \leq\frac{e^{2r}}{r!}=\mathcal{O}(1). $$ From this we obtain $$ \#End(T_n)\leq Cn! n^3\max_m\frac{m^{n-m}}{(n-m)!} \leq Cn! n^3\max_m\frac{m^{n-m}}{(n-m)!}. $$ Put $m=\alpha n$. Then $$ \frac{m^{n-m}}{(n-m)!}\leq \left(\frac{e\alpha}{1-\alpha}\right)^{(1-\alpha)n}. $$ The maximum of $\left(\frac{e\alpha}{1-\alpha}\right)^{(1-\alpha)}$ is about $1.763$, thus for large $n$ we have $\#End(T_n)\leq 1.77^n n!$, which is a lot smaller than $n^n\sim \frac{e^n n!}{\sqrt{2\pi n}}$.

Looking at the exponential series, we have $\frac{r^k}{k!}\leq e^r$. Hence we have $$ \frac{r^{m-k-r}}{(m-2k-r)!r!k!} = \frac{1}{r!}\frac{r^k}{k!}\frac{r^{m-2k-r}}{(m-2k-r)!} \leq\frac{e^{2r}}{r!}=\mathcal{O}(1). $$ From this we obtain $$ \#End(T_n)\leq Cn! n^3\max_m\frac{m^{n-m}}{(n-m)!} \leq Cn! n^3\max_m\frac{m^{n-m}}{(n-m)!}. $$ Put $m=\alpha n$. Then $$ \frac{m^{n-m}}{(n-m)!}\leq \left(\frac{e\alpha}{1-\alpha}\right)^{(1-\alpha)n}. $$ The maximum of $\left(\frac{e\alpha}{1-\alpha}\right)^{(1-\alpha)}$ is about $1.32$, thus for large $n$ we have $\#End(T_n)\leq 1.33^n n!$, which is a lot smaller than $n^n\sim e^n n!\sqrt{2\pi n}$.

Looking at the exponential series, we have $\frac{r^k}{k!}\leq e^r$. Hence we have $$ \frac{r^{m-k-r}}{(m-2k-r)!r!k!} = \frac{1}{r!}\frac{r^k}{k!}\frac{r^{m-2k-r}}{(m-2k-r)!} \leq\frac{e^{2r}}{r!}=\mathcal{O}(1). $$ From this we obtain $$ \#End(T_n)\leq Cn! n^3\max_m\frac{m^{n-m}}{(n-m)!} \leq Cn! n^3\max_m\frac{m^{n-m}}{(n-m)!}. $$ Put $m=\alpha n$. Then $$ \frac{m^{n-m}}{(n-m)!}\leq \left(\frac{e\alpha}{1-\alpha}\right)^{(1-\alpha)n}. $$ The maximum of $\left(\frac{e\alpha}{1-\alpha}\right)^{(1-\alpha)}$ is about $1.783$, thus for large $n$ we have $\#End(T_n)\leq 1.79^n n!$, which is a lot smaller than $n^n\sim e^n n!\sqrt{2\pi n}$.