As an exercise I'm trying to verify that for $X=\Bbb{P}_k^n$, where $k$ is an algebraically closed field, we have $$\operatorname{td}(X)=\left(\frac{\epsilon}{1-e^{-\epsilon}}\right)^{n+1},$$ where $\epsilon\in A(X)$ is the class of a hyperplane in the Chow ring of $X$. By definition we have $\operatorname{td}(X):=\operatorname{td}(\mathcal{T}_X)$, where $\mathcal{T}_X$ is the tangent sheaf on $X$, and in turn for any vector bundle $\mathcal{E}$ on $X$ we have by definition $$\operatorname{td}(\mathcal{E}):=\prod_{i=1}^r\frac{\alpha_i}{1-e^{-\alpha_i}},$$ where $r=\operatorname{rk}(\mathcal{E})$ and the $\alpha_i$ are such that $c_t(\mathcal{E})=\prod_{i=1}^r(1+\alpha_it)$.
I've managed to show that $\operatorname{rk}(\mathcal{T}_X)=n$ and $c_t(\mathcal{T}_X)=\prod_{i=1}^{n+1}(1+\epsilon t)$. Tempting as it may be to say that $\alpha_i=\epsilon$ for all $i$ and be done, unfortunately the indices don't match up as $n\neq n+1$.
I haven't been able to fix this problem: I have shown that $\epsilon^{n+1}=0$ and so we may write $$c_t(\mathcal{T}_X)=\prod_{i=1}^{n+1}(1+\epsilon t)=\sum_{i=0}^{n+1}\binom{n+1}{i}\epsilon^it^i=\sum_{i=0}^n\binom{n+1}{i}\epsilon^it^i,$$ from which it follows that $\prod_{i=1}^n\alpha_i=\binom{n+1}{n}\epsilon^n=(n+1)\epsilon^n$, and hence $$\operatorname{td}(\mathcal{T}_X)=\prod_{i=1}^r\frac{\alpha_i}{1-e^{-\alpha_i}}=\prod_{i=1}^n\alpha_i\cdot\prod_{i=1}^n\frac{1}{1-e^{-\alpha_i}}=(n+1)\epsilon^n\cdot\prod_{i=1}^n\frac{1}{1-e^{-\alpha_i}}.$$ At this point I'm stuck; clearly $\prod_{i=1}^n\frac{1}{1-e^{-\alpha_i}}$ is a symmetric polynomial in the $\alpha_i$ and hence a polynomial in the coefficients of $c_t(\mathcal{T}_X)$, i.e. in the terms $\binom{n+1}{i}\epsilon^i$ for $i\in\{0,\ldots,n\}$. But how to find this polynomial explicitly?
To compute $td(\mathbb P^n)$, one can use that $td$ is multiplicative in exact sequences, and look at the Euler sequence on $\mathbb P^n$: $$0\to \mathscr O\to \mathscr O(1)^{\oplus (n+1)}\to T_{\mathbb P^n}\to 0.$$ Knowing that $td(\mathscr O)=1$, we get $$td(T_{\mathbb P^n})=td(\mathscr O(1)^{\oplus (n+1)})=td(\mathscr O(1))^{n+1}=\Bigl(\frac{\epsilon}{1-e^{-\epsilon}}\Bigr)^{n+1},$$ where $c_t(\mathscr O(1))=1+\epsilon t$.
For the first equality, it is enough to notice that the tangent bundle $T_{\mathbb P^n}$ and the vector bundle $\mathscr O(1)^{\oplus (n+1)}$ have the same Chern classes, hence the same Todd class. But for the second equality one really needs the multiplicativity of $td$.